{"id":543612,"date":"2021-09-30T06:55:52","date_gmt":"2021-09-30T06:55:52","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=543612"},"modified":"2021-10-01T09:08:57","modified_gmt":"2021-10-01T09:08:57","slug":"rd-sharma-solutions-for-class-11-maths-chapter-20-geometric-progressions","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-20-geometric-progressions\/","title":{"rendered":"RD Sharma Solutions for Class 11 Maths Chapter 20\u2013Geometric Progressions"},"content":{"rendered":"\n

Class 11: Maths Chapter 20 solutions. Complete Class 11 Maths Chapter 20 Notes.<\/p>\n\n\n\n

RD Sharma Solutions for Class 11 Maths Chapter 20\u2013Geometric Progressions<\/h2>\n\n\n\n

RD Sharma 11th Maths Chapter 20, Class 11 Maths Chapter 20 solutions<\/p>\n\n\n\n

EXERCISE 20.1 PAGE NO: 20.9<\/p>\n\n\n\n

1. Show that each one of the following progressions is a G.P. Also, find the common ratio in each case:<\/strong><\/p>\n\n\n\n

(i) 4, -2, 1, -1\/2, \u2026.<\/strong><\/p>\n\n\n\n

(ii) -2\/3, -6, -54, \u2026.<\/strong><\/p>\n\n\n\n

(iii) a, 3a2<\/sup>\/4, 9a3<\/sup>\/16, \u2026.<\/strong><\/p>\n\n\n\n

(iv) \u00bd, 1\/3, 2\/9, 4\/27, \u2026<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) <\/strong>4, -2, 1, -1\/2, \u2026.<\/p>\n\n\n\n

Let a = 4, b = -2, c = 1<\/p>\n\n\n\n

In GP,<\/p>\n\n\n\n

b2 <\/sup>= ac<\/p>\n\n\n\n

(-2)2<\/sup> = 4(1)<\/p>\n\n\n\n

4 = 4<\/p>\n\n\n\n

So, the Common ratio = r = -2\/4 = -1\/2<\/p>\n\n\n\n

(ii) <\/strong>-2\/3, -6, -54, \u2026.<\/p>\n\n\n\n

Let a = -2\/3, b = -6, c = -54<\/p>\n\n\n\n

In GP,<\/p>\n\n\n\n

b2 <\/sup>= ac<\/p>\n\n\n\n

(-6)2<\/sup> = -2\/3 \u00d7 (-54)<\/p>\n\n\n\n

36 = 36<\/p>\n\n\n\n

So, the Common ratio = r = -6\/(-2\/3) = -6 \u00d7 3\/-2 = 9<\/p>\n\n\n\n

(iii) <\/strong>a, 3a2<\/sup>\/4, 9a3<\/sup>\/16, \u2026.<\/p>\n\n\n\n

Let a = a, b = 3a2<\/sup>\/4, c = 9a3<\/sup>\/16<\/p>\n\n\n\n

In GP,<\/p>\n\n\n\n

b2 <\/sup>= ac<\/p>\n\n\n\n

(3a2<\/sup>\/4)2<\/sup> = 9a3<\/sup>\/16 \u00d7 a<\/p>\n\n\n\n

9a4<\/sup>\/4 = 9a4<\/sup>\/16<\/p>\n\n\n\n

So, the Common ratio = r = (3a2<\/sup>\/4)\/a = 3a2<\/sup>\/4a = 3a\/4<\/p>\n\n\n\n

(iv) <\/strong>\u00bd, 1\/3, 2\/9, 4\/27, \u2026<\/p>\n\n\n\n

Let a = 1\/2, b = 1\/3, c = 2\/9<\/p>\n\n\n\n

In GP,<\/p>\n\n\n\n

b2 <\/sup>= ac<\/p>\n\n\n\n

(1\/3)2<\/sup> = 1\/2 \u00d7 (2\/9)<\/p>\n\n\n\n

1\/9 = 1\/9<\/p>\n\n\n\n

So, the Common ratio = r = (1\/3)\/(1\/2) = (1\/3) \u00d7 2 = 2\/3<\/p>\n\n\n\n

2. Show that the sequence defined by an<\/sub> = 2\/3n<\/sup>, n \u2208 N is a G.P.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

an<\/sub> = 2\/3n<\/sup><\/p>\n\n\n\n

Let us consider n = 1, 2, 3, 4, \u2026 since n is a natural number.<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

a1<\/sub> = 2\/3<\/p>\n\n\n\n

a2<\/sub> = 2\/32<\/sup> = 2\/9<\/p>\n\n\n\n

a3<\/sub> = 2\/33<\/sup> = 2\/27<\/p>\n\n\n\n

a4<\/sub> = 2\/34<\/sup> = 2\/81<\/p>\n\n\n\n

In GP,<\/p>\n\n\n\n

a3<\/sub>\/a2<\/sub> = (2\/27) \/ (2\/9)<\/p>\n\n\n\n

= 2\/27 \u00d7 9\/2<\/p>\n\n\n\n

= 1\/3<\/p>\n\n\n\n

a2<\/sub>\/a1<\/sub> = (2\/9) \/ (2\/3)<\/p>\n\n\n\n

= 2\/9 \u00d7 3\/2<\/p>\n\n\n\n

= 1\/3<\/p>\n\n\n\n

\u2234 Common ratio of consecutive term is 1\/3. Hence n \u2208 N is a G.P.<\/p>\n\n\n\n

3. Find:<\/strong><\/p>\n\n\n\n

(i) the ninth term of the G.P. 1, 4, 16, 64, \u2026.<\/strong><\/p>\n\n\n\n

(ii) the 10th<\/sup> term of the G.P. -3\/4, \u00bd, -1\/3, 2\/9, \u2026.<\/strong><\/p>\n\n\n\n

(iii) the 8th<\/sup> term of the G.P. 0.3, 0.06, 0.012, \u2026.<\/strong><\/p>\n\n\n\n

(iv) the 12th<\/sup> term of the G.P. 1\/a3<\/sup>x3<\/sup> , ax, a5<\/sup>x5<\/sup>, \u2026.<\/strong><\/p>\n\n\n\n

(v) nth term of the G.P. \u221a3, 1\/\u221a3, 1\/3\u221a3, \u2026<\/strong><\/p>\n\n\n\n

(vi) the 10th<\/sup> term of the G.P. \u221a2, 1\/\u221a2, 1\/2\u221a2, \u2026.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) <\/strong>the ninth term of the G.P. 1, 4, 16, 64, \u2026.<\/p>\n\n\n\n

We know that,<\/p>\n\n\n\n

t1<\/sub> = a = 1, r = t2<\/sub>\/t1<\/sub> = 4\/1 = 4<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Tn<\/sub> = arn-1<\/sup><\/p>\n\n\n\n

T9<\/sub> = 1 (4)9-1<\/sup><\/p>\n\n\n\n

= 1 (4)8<\/sup><\/p>\n\n\n\n

= 48<\/sup><\/p>\n\n\n\n

(ii) <\/strong>the 10th<\/sup> term of the G.P. -3\/4, \u00bd, -1\/3, 2\/9, \u2026.<\/p>\n\n\n\n

We know that,<\/p>\n\n\n\n

t1<\/sub> = a = -3\/4, r = t2<\/sub>\/t1<\/sub> = (1\/2) \/ (-3\/4) = \u00bd \u00d7 -4\/3 = -2\/3<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Tn<\/sub> = arn-1<\/sup><\/p>\n\n\n\n

T10<\/sub> = -3\/4 (-2\/3)10-1<\/sup><\/p>\n\n\n\n

= -3\/4 (-2\/3)9<\/sup><\/p>\n\n\n\n

= \u00bd (2\/3)8<\/sup><\/p>\n\n\n\n

(iii) <\/strong>the 8th<\/sup> term of the G.P., 0.3, 0.06, 0.012, \u2026.<\/p>\n\n\n\n

We know that,<\/p>\n\n\n\n

t1<\/sub> = a = 0.3, r = t2<\/sub>\/t1<\/sub> = 0.06\/0.3 = 0.2<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Tn<\/sub> = arn-1<\/sup><\/p>\n\n\n\n

T8<\/sub> = 0.3 (0.2)8-1<\/sup><\/p>\n\n\n\n

= 0.3 (0.2)7<\/sup><\/p>\n\n\n\n

(iv) <\/strong>the 12th<\/sup> term of the G.P. 1\/a3<\/sup>x3<\/sup> , ax, a5<\/sup>x5<\/sup>, \u2026.<\/p>\n\n\n\n

We know that,<\/p>\n\n\n\n

t1<\/sub> = a = 1\/a3<\/sup>x3<\/sup>, r = t2<\/sub>\/t1<\/sub> = ax\/(1\/a3<\/sup>x3<\/sup>) = ax (a3<\/sup>x3<\/sup>) = a4<\/sup>x4<\/sup><\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Tn<\/sub> = arn-1<\/sup><\/p>\n\n\n\n

T12<\/sub> = 1\/a3<\/sup>x3<\/sup> (a4<\/sup>x4<\/sup>)12-1<\/sup><\/p>\n\n\n\n

= 1\/a3<\/sup>x3<\/sup> (a4<\/sup>x4<\/sup>)11<\/sup><\/p>\n\n\n\n

= (ax)41<\/sup><\/p>\n\n\n\n

(v) <\/strong>nth term of the G.P. \u221a3, 1\/\u221a3, 1\/3\u221a3, \u2026<\/p>\n\n\n\n

We know that,<\/p>\n\n\n\n

t1<\/sub> = a = \u221a3, r = t2<\/sub>\/t1<\/sub> = (1\/\u221a3)\/\u221a3 = 1\/(\u221a3\u00d7\u221a3) = 1\/3<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Tn<\/sub> = arn-1<\/sup><\/p>\n\n\n\n

Tn<\/sub> = \u221a3 (1\/3)n-1<\/sup><\/p>\n\n\n\n

(vi) <\/strong>the 10th<\/sup> term of the G.P. \u221a2, 1\/\u221a2, 1\/2\u221a2, \u2026.<\/p>\n\n\n\n

We know that,<\/p>\n\n\n\n

t1<\/sub> = a = \u221a2, r = t2<\/sub>\/t1<\/sub> = (1\/\u221a2)\/\u221a2 = 1\/(\u221a2\u00d7\u221a2) = 1\/2<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Tn<\/sub> = arn-1<\/sup><\/p>\n\n\n\n

T10<\/sub> = \u221a2 (1\/2)10-1<\/sup><\/p>\n\n\n\n

= \u221a2 (1\/2)9<\/sup><\/p>\n\n\n\n

= 1\/\u221a2 (1\/2)8<\/sup><\/p>\n\n\n\n

4. Find the 4th<\/sup> term from the end of the G.P. 2\/27, 2\/9, 2\/3, \u2026., 162.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The nth term from the end is given by:<\/p>\n\n\n\n

an<\/sub> = l (1\/r)n-1<\/sup> where, l is the last term, r is the common ratio, n is the nth term<\/p>\n\n\n\n

Given: last term, l = 162<\/p>\n\n\n\n

r = t2<\/sub>\/t1<\/sub> = (2\/9) \/ (2\/27)<\/p>\n\n\n\n

= 2\/9 \u00d7 27\/2<\/p>\n\n\n\n

= 3<\/p>\n\n\n\n

n = 4<\/p>\n\n\n\n

So, an<\/sub> = l (1\/r)n-1<\/sup><\/p>\n\n\n\n

a4<\/sub> = 162 (1\/3)4-1<\/sup><\/p>\n\n\n\n

= 162 (1\/3)3<\/sup><\/p>\n\n\n\n

= 162 \u00d7 1\/27<\/p>\n\n\n\n

= 6<\/p>\n\n\n\n

\u2234 4th<\/sup> term from last is 6.<\/p>\n\n\n\n

5. Which term of the progression 0.004, 0.02, 0.1, \u2026. is 12.5?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Tn<\/sub> = arn-1<\/sup><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

a = 0.004<\/p>\n\n\n\n

r = t2<\/sub>\/t1<\/sub> = (0.02\/0.004)<\/p>\n\n\n\n

= 5<\/p>\n\n\n\n

Tn<\/sub> = 12.5<\/p>\n\n\n\n

n = ?<\/p>\n\n\n\n

So, Tn<\/sub> = arn-1<\/sup><\/p>\n\n\n\n

12.5 = (0.004) (5)n-1<\/sup><\/p>\n\n\n\n

12.5\/0.004 = 5n-1<\/sup><\/p>\n\n\n\n

3000 = 5n-1<\/sup><\/p>\n\n\n\n

55<\/sup> = 5n-1<\/sup><\/p>\n\n\n\n

5 = n-1<\/p>\n\n\n\n

n = 5 + 1<\/p>\n\n\n\n

= 6<\/p>\n\n\n\n

\u2234 6th<\/sup> term of the progression 0.004, 0.02, 0.1, \u2026. is 12.5.<\/p>\n\n\n\n

6. Which term of the G.P.:<\/strong><\/p>\n\n\n\n

(i) \u221a2, 1\/\u221a2, 1\/2\u221a2, 1\/4\u221a2, \u2026 is 1\/512\u221a2 ?<\/strong><\/p>\n\n\n\n

(ii) 2, 2\u221a2, 4, \u2026 is 128 ?<\/strong><\/p>\n\n\n\n

(iii) \u221a3, 3, 3\u221a3, \u2026 is 729 ?<\/strong><\/p>\n\n\n\n

(iv) 1\/3, 1\/9, 1\/27\u2026 is 1\/19683 ?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) <\/strong>\u221a2, 1\/\u221a2, 1\/2\u221a2, 1\/4\u221a2, \u2026 is 1\/512\u221a2 ?<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Tn<\/sub> = arn-1<\/sup><\/p>\n\n\n\n

a = \u221a2<\/p>\n\n\n\n

r = t2<\/sub>\/t1<\/sub> = (1\/\u221a2) \/ (\u221a2)<\/p>\n\n\n\n

= 1\/2<\/p>\n\n\n\n

Tn<\/sub> = 1\/512\u221a2<\/p>\n\n\n\n

n = ?<\/p>\n\n\n\n

Tn<\/sub> = arn-1<\/sup><\/p>\n\n\n\n

1\/512\u221a2 = (\u221a2) (1\/2)n-1<\/sup><\/p>\n\n\n\n

1\/512\u221a2\u00d7\u221a2 = (1\/2)n-1<\/sup><\/p>\n\n\n\n

1\/512\u00d72 = (1\/2)n-1<\/sup><\/p>\n\n\n\n

1\/1024 = (1\/2)n-1<\/sup><\/p>\n\n\n\n

(1\/2)10<\/sup> = (1\/2)n-1<\/sup><\/p>\n\n\n\n

10 = n \u2013 1<\/p>\n\n\n\n

n = 10 + 1<\/p>\n\n\n\n

= 11<\/p>\n\n\n\n

\u2234 11th<\/sup> term of the G.P is 1\/512\u221a2<\/p>\n\n\n\n

(ii) <\/strong>2, 2\u221a2, 4, \u2026 is 128 ?<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Tn<\/sub> = arn-1<\/sup><\/p>\n\n\n\n

a = 2<\/p>\n\n\n\n

r = t2<\/sub>\/t1<\/sub> = (2\u221a2\/2)<\/p>\n\n\n\n

= \u221a2<\/p>\n\n\n\n

Tn<\/sub> = 128<\/p>\n\n\n\n

n = ?<\/p>\n\n\n\n

Tn<\/sub> = arn-1<\/sup><\/p>\n\n\n\n

128 = 2 (\u221a2)n-1<\/sup><\/p>\n\n\n\n

128\/2 = (\u221a2)n-1<\/sup><\/p>\n\n\n\n

64 = (\u221a2)n-1<\/sup><\/p>\n\n\n\n

26<\/sup> = (\u221a2)n-1<\/sup><\/p>\n\n\n\n

12 = n \u2013 1<\/p>\n\n\n\n

n = 12 + 1<\/p>\n\n\n\n

= 13<\/p>\n\n\n\n

\u2234 13th<\/sup> term of the G.P is 128<\/p>\n\n\n\n

(iii) <\/strong>\u221a3, 3, 3\u221a3, \u2026 is 729 ?<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Tn<\/sub> = arn-1<\/sup><\/p>\n\n\n\n

a = \u221a3<\/p>\n\n\n\n

r = t2<\/sub>\/t1<\/sub> = (3\/\u221a3)<\/p>\n\n\n\n

= \u221a3<\/p>\n\n\n\n

Tn<\/sub> = 729<\/p>\n\n\n\n

n = ?<\/p>\n\n\n\n

Tn<\/sub> = arn-1<\/sup><\/p>\n\n\n\n

729 = \u221a3 (\u221a3)n-1<\/sup><\/p>\n\n\n\n

729 = (\u221a3)n<\/sup><\/p>\n\n\n\n

36<\/sup> = (\u221a3)n<\/sup><\/p>\n\n\n\n

(\u221a3)12<\/sup> = (\u221a3)n<\/sup><\/p>\n\n\n\n

n = 12<\/p>\n\n\n\n

\u2234 12th<\/sup> term of the G.P is 729<\/p>\n\n\n\n

(iv) <\/strong>1\/3, 1\/9, 1\/27\u2026 is 1\/19683 ?<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Tn<\/sub> = arn-1<\/sup><\/p>\n\n\n\n

a = 1\/3<\/p>\n\n\n\n

r = t2<\/sub>\/t1<\/sub> = (1\/9) \/ (1\/3)<\/p>\n\n\n\n

= 1\/9 \u00d7 3\/1<\/p>\n\n\n\n

= 1\/3<\/p>\n\n\n\n

Tn<\/sub> = 1\/19683<\/p>\n\n\n\n

n = ?<\/p>\n\n\n\n

Tn<\/sub> = arn-1<\/sup><\/p>\n\n\n\n

1\/19683 = (1\/3) (1\/3)n-1<\/sup><\/p>\n\n\n\n

1\/19683 = (1\/3)n<\/sup><\/p>\n\n\n\n

(1\/3)9<\/sup> = (1\/3)n<\/sup><\/p>\n\n\n\n

n = 9<\/p>\n\n\n\n

\u2234 9th<\/sup> term of the G.P is 1\/19683<\/p>\n\n\n\n

7. Which term of the progression 18, -12, 8, \u2026 is 512\/729 ?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Tn<\/sub> = arn-1<\/sup><\/p>\n\n\n\n

a = 18<\/p>\n\n\n\n

r = t2<\/sub>\/t1<\/sub> = (-12\/18)<\/p>\n\n\n\n

= -2\/3<\/p>\n\n\n\n

Tn<\/sub> = 512\/729<\/p>\n\n\n\n

n = ?<\/p>\n\n\n\n

Tn<\/sub> = arn-1<\/sup><\/p>\n\n\n\n

512\/729 = 18 (-2\/3)n-1<\/sup><\/p>\n\n\n\n

29<\/sup>\/(729 \u00d7 18) = (-2\/3)n-1<\/sup><\/p>\n\n\n\n

29<\/sup>\/36 \u00d7 1\/2\u00d732<\/sup> = (-2\/3)n-1<\/sup><\/p>\n\n\n\n

(2\/3)8<\/sup> = (-1)n-1<\/sup> (2\/3)n-1<\/sup><\/p>\n\n\n\n

8 = n \u2013 1<\/p>\n\n\n\n

n = 8 + 1<\/p>\n\n\n\n

= 9<\/p>\n\n\n\n

\u2234 9th<\/sup> term of the Progression is 512\/729<\/p>\n\n\n\n

8. Find the 4th term from the end of the G.P. \u00bd, 1\/6, 1\/18, 1\/54, \u2026 , 1\/4374<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The nth term from the end is given by:<\/p>\n\n\n\n

an<\/sub> = l (1\/r)n-1<\/sup> where, l is the last term, r is the common ratio, n is the nth term<\/p>\n\n\n\n

Given: last term, l = 1\/4374<\/p>\n\n\n\n

r = t2<\/sub>\/t1<\/sub> = (1\/6) \/ (1\/2)<\/p>\n\n\n\n

= 1\/6 \u00d7 2\/1<\/p>\n\n\n\n

= 1\/3<\/p>\n\n\n\n

n = 4<\/p>\n\n\n\n

So, an<\/sub> = l (1\/r)n-1<\/sup><\/p>\n\n\n\n

a4<\/sub> = 1\/4374 (1\/(1\/3))4-1<\/sup><\/p>\n\n\n\n

= 1\/4374 (3\/1)3<\/sup><\/p>\n\n\n\n

= 1\/4374 \u00d7 33<\/sup><\/p>\n\n\n\n

= 1\/4374 \u00d7 27<\/p>\n\n\n\n

= 1\/162<\/p>\n\n\n\n

\u2234 4th<\/sup> term from last is 1\/162.<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 20.2 PAGE NO: 20.16<\/p>\n\n\n\n

1. Find three numbers in G.P. whose sum is 65 and whose product is 3375.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let the three numbers be a\/r, a, ar<\/p>\n\n\n\n

So, according to the question<\/p>\n\n\n\n

a\/r + a + ar = 65 \u2026 equation (1)<\/p>\n\n\n\n

a\/r \u00d7 a \u00d7 ar = 3375 \u2026 equation (2)<\/p>\n\n\n\n

From equation (2) we get,<\/p>\n\n\n\n

a3<\/sup> = 3375<\/p>\n\n\n\n

a = 15.<\/p>\n\n\n\n

From equation (1) we get,<\/p>\n\n\n\n

(a + ar + ar2<\/sup>)\/r = 65<\/p>\n\n\n\n

a + ar + ar2<\/sup> = 65r \u2026 equation (3)<\/p>\n\n\n\n

Substituting a = 15 in equation (3) we get<\/p>\n\n\n\n

15 + 15r + 15r2<\/sup> = 65r<\/p>\n\n\n\n

15r2<\/sup> \u2013 50r + 15 = 0\u2026 equation (4)<\/p>\n\n\n\n

Dividing equation (4) by 5 we get<\/p>\n\n\n\n

3r2<\/sup> \u2013 10r + 3 = 0<\/p>\n\n\n\n

3r2<\/sup> \u2013 9r \u2013 r + 3 = 0<\/p>\n\n\n\n

3r(r \u2013 3) \u2013 1(r \u2013 3) = 0<\/p>\n\n\n\n

r = 3 or r = 1\/3<\/p>\n\n\n\n

Now, the equation will be<\/p>\n\n\n\n

15\/3, 15, 15\u00d73 or<\/p>\n\n\n\n

15\/(1\/3), 15, 15\u00d71\/3<\/p>\n\n\n\n

So the terms are 5, 15, 45 or 45, 15, 5<\/p>\n\n\n\n

\u2234 The three numbers are 5, 15, 45.<\/p>\n\n\n\n

2. Find three number in G.P. whose sum is 38 and their product is 1728.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let the three numbers be a\/r, a, ar<\/p>\n\n\n\n

So, according to the question<\/p>\n\n\n\n

a\/r + a + ar = 38 \u2026 equation (1)<\/p>\n\n\n\n

a\/r \u00d7 a \u00d7 ar = 1728 \u2026 equation (2)<\/p>\n\n\n\n

From equation (2) we get,<\/p>\n\n\n\n

a3<\/sup> = 1728<\/p>\n\n\n\n

a = 12.<\/p>\n\n\n\n

From equation (1) we get,<\/p>\n\n\n\n

(a + ar + ar2<\/sup>)\/r = 38<\/p>\n\n\n\n

a + ar + ar2<\/sup> = 38r \u2026 equation (3)<\/p>\n\n\n\n

Substituting a = 12 in equation (3) we get<\/p>\n\n\n\n

12 + 12r + 12r2<\/sup> = 38r<\/p>\n\n\n\n

12r2<\/sup> \u2013 26r + 12 = 0\u2026 equation (4)<\/p>\n\n\n\n

Dividing equation (4) by 2 we get<\/p>\n\n\n\n

6r2<\/sup> \u2013 13r + 6 = 0<\/p>\n\n\n\n

6r2<\/sup> \u2013 9r \u2013 4r + 6 = 0<\/p>\n\n\n\n

3r(3r \u2013 3) \u2013 2(3r \u2013 3) = 0<\/p>\n\n\n\n

r = 3\/2 or r = 2\/3<\/p>\n\n\n\n

Now the equation will be<\/p>\n\n\n\n

12\/(3\/2) = 8 or<\/p>\n\n\n\n

12\/(2\/3) = 18<\/p>\n\n\n\n

So the terms are 8, 12, 18<\/p>\n\n\n\n

\u2234 The three numbers are 8, 12, 18<\/p>\n\n\n\n

3. The sum of first three terms of a G.P. is 13\/12, and their product is \u2013 1. Find the G.P.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let the three numbers be a\/r, a, ar<\/p>\n\n\n\n

So, according to the question<\/p>\n\n\n\n

a\/r + a + ar = 13\/12 \u2026 equation (1)<\/p>\n\n\n\n

a\/r \u00d7 a \u00d7 ar = -1 \u2026 equation (2)<\/p>\n\n\n\n

From equation (2) we get,<\/p>\n\n\n\n

a3<\/sup> = -1<\/p>\n\n\n\n

a = -1<\/p>\n\n\n\n

From equation (1) we get,<\/p>\n\n\n\n

(a + ar + ar2<\/sup>)\/r = 13\/12<\/p>\n\n\n\n

12a + 12ar + 12ar2<\/sup> = 13r \u2026 equation (3)<\/p>\n\n\n\n

Substituting a = \u2013 1 in equation (3) we get<\/p>\n\n\n\n

12( \u2013 1) + 12( \u2013 1)r + 12( \u2013 1)r2<\/sup> = 13r<\/p>\n\n\n\n

12r2<\/sup> + 25r + 12 = 0<\/p>\n\n\n\n

12r2<\/sup> + 16r + 9r + 12 = 0\u2026 equation (4)<\/p>\n\n\n\n

4r (3r + 4) + 3(3r + 4) = 0<\/p>\n\n\n\n

r = -3\/4 or r = -4\/3<\/p>\n\n\n\n

Now the equation will be<\/p>\n\n\n\n

-1\/(-3\/4), -1, -1\u00d7-3\/4 or -1\/(-4\/3), -1, -1\u00d7-4\/3<\/p>\n\n\n\n

4\/3, -1, \u00be or \u00be, -1, 4\/3<\/p>\n\n\n\n

\u2234 The three numbers are 4\/3, -1, \u00be or \u00be, -1, 4\/3<\/p>\n\n\n\n

4. The product of three numbers in G.P. is 125 and the sum of their products taken in pairs is 87 \u00bd . Find them.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let the three numbers be a\/r, a, ar<\/p>\n\n\n\n

So, according to the question<\/p>\n\n\n\n

a\/r \u00d7 a \u00d7 ar = 125 \u2026 equation (1)<\/p>\n\n\n\n

From equation (1) we get,<\/p>\n\n\n\n

a3<\/sup> = 125<\/p>\n\n\n\n

a = 5<\/p>\n\n\n\n

a\/r \u00d7 a + a \u00d7 ar + ar \u00d7 a\/r = 87 \u00bd<\/p>\n\n\n\n

a\/r \u00d7 a + a \u00d7 ar + ar \u00d7 a\/r = 195\/2<\/p>\n\n\n\n

a2<\/sup>\/r + a2<\/sup>r + a2<\/sup> = 195\/2<\/p>\n\n\n\n

a2<\/sup> (1\/r + r + 1) = 195\/2<\/p>\n\n\n\n

Substituting a = 5 in above equation we get,<\/p>\n\n\n\n

52<\/sup> [(1+r2<\/sup>+r)\/r] = 195\/2<\/p>\n\n\n\n

1+r2<\/sup>+r = (195r\/2\u00d725)<\/p>\n\n\n\n

2(1+r2<\/sup>+r) = 39r\/5<\/p>\n\n\n\n

10 + 10r2<\/sup> + 10r = 39r<\/p>\n\n\n\n

10r2<\/sup> \u2013 29r + 10 = 0<\/p>\n\n\n\n

10r2<\/sup> \u2013 25r \u2013 4r + 10 = 0<\/p>\n\n\n\n

5r(2r-5) \u2013 2(2r-5) = 0<\/p>\n\n\n\n

r = 5\/2, 2\/5<\/p>\n\n\n\n

So G.P is 10, 5, 5\/2 or 5\/2, 5, 10<\/p>\n\n\n\n

\u2234 The three numbers are 10, 5, 5\/2 or 5\/2, 5, 10<\/p>\n\n\n\n

5. The sum of the first three terms of a G.P. is 39\/10, and their product is 1. Find the common ratio and the terms.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let the three numbers be a\/r, a, ar<\/p>\n\n\n\n

So, according to the question<\/p>\n\n\n\n

a\/r + a + ar = 39\/10 \u2026 equation (1)<\/p>\n\n\n\n

a\/r \u00d7 a \u00d7 ar = 1 \u2026 equation (2)<\/p>\n\n\n\n

From equation (2) we get,<\/p>\n\n\n\n

a3<\/sup> = 1<\/p>\n\n\n\n

a = 1<\/p>\n\n\n\n

From equation (1) we get,<\/p>\n\n\n\n

(a + ar + ar2<\/sup>)\/r = 39\/10<\/p>\n\n\n\n

10a + 10ar + 10ar2<\/sup> = 39r \u2026 equation (3)<\/p>\n\n\n\n

Substituting a = 1 in 3 we get<\/p>\n\n\n\n

10(1) + 10(1)r + 10(1)r2<\/sup> = 39r<\/p>\n\n\n\n

10r2<\/sup> \u2013 29r + 10 = 0<\/p>\n\n\n\n

10r2<\/sup> \u2013 25r \u2013 4r + 10 = 0\u2026 equation (4)<\/p>\n\n\n\n

5r(2r \u2013 5) \u2013 2(2r \u2013 5) = 0<\/p>\n\n\n\n

r = 2\/5 or 5\/2<\/p>\n\n\n\n

so now the equation will be,<\/p>\n\n\n\n

1\/(2\/5), 1, 1\u00d72\/5 or 1\/(5\/2), 1, 1\u00d75\/2<\/p>\n\n\n\n

5\/2, 1, 2\/5 or 2\/5, 1, 5\/2<\/p>\n\n\n\n

\u2234 The three numbers are 2\/5, 1, 5\/2<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 20.3 PAGE NO: 20.27<\/p>\n\n\n\n

1. Find the sum of the following geometric progressions:<\/strong><\/p>\n\n\n\n

(i) 2, 6, 18, \u2026 to 7 terms<\/strong><\/p>\n\n\n\n

(ii) 1, 3, 9, 27, \u2026 to 8 terms<\/strong><\/p>\n\n\n\n

(iii) 1, -1\/2, \u00bc, -1\/8, \u2026<\/strong><\/p>\n\n\n\n

(iv) (a2<\/sup> \u2013 b2<\/sup>), (a \u2013 b), (a-b)\/(a+b), \u2026 to n terms<\/strong><\/p>\n\n\n\n

(v) 4, 2, 1, \u00bd \u2026 to 10 terms<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) <\/strong>2, 6, 18, \u2026 to 7 terms<\/p>\n\n\n\n

We know that, sum of GP for n terms = a(rn<\/sup> \u2013 1)\/(r \u2013 1)<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

a = 2, r = t2<\/sub>\/t1<\/sub> = 6\/2 = 3, n = 7<\/p>\n\n\n\n

Now let us substitute the values in<\/p>\n\n\n\n

a(rn<\/sup> \u2013 1)\/(r \u2013 1) = 2 (37<\/sup> \u2013 1)\/(3-1)<\/p>\n\n\n\n

= 2 (37<\/sup> \u2013 1)\/2<\/p>\n\n\n\n

= 37<\/sup> \u2013 1<\/p>\n\n\n\n

= 2187 \u2013 1<\/p>\n\n\n\n

= 2186<\/p>\n\n\n\n

(ii) <\/strong>1, 3, 9, 27, \u2026 to 8 terms<\/p>\n\n\n\n

We know that, sum of GP for n terms = a(rn<\/sup> \u2013 1)\/(r \u2013 1)<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

a = 1, r = t2<\/sub>\/t1<\/sub> = 3\/1 = 3, n = 8<\/p>\n\n\n\n

Now let us substitute the values in<\/p>\n\n\n\n

a(rn<\/sup> \u2013 1)\/(r \u2013 1) = 1 (38<\/sup> \u2013 1)\/(3-1)<\/p>\n\n\n\n

= (38<\/sup> \u2013 1)\/2<\/p>\n\n\n\n

= (6561 \u2013 1)\/2<\/p>\n\n\n\n

= 6560\/2<\/p>\n\n\n\n

= 3280<\/p>\n\n\n\n

(iii) <\/strong>1, -1\/2, \u00bc, -1\/8, \u2026<\/p>\n\n\n\n

We know that, sum of GP for infinity = a\/(1 \u2013 r)<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

a = 1, r = t2<\/sub>\/t1<\/sub> = (-1\/2)\/1 = -1\/2<\/p>\n\n\n\n

Now let us substitute the values in<\/p>\n\n\n\n

a\/(1 \u2013 r) = 1\/(1 \u2013 (-1\/2))<\/p>\n\n\n\n

= 1\/(1 + 1\/2)<\/p>\n\n\n\n

= 1\/((2+1)\/2)<\/p>\n\n\n\n

= 1\/(3\/2)<\/p>\n\n\n\n

= 2\/3<\/p>\n\n\n\n

(iv) <\/strong>(a2<\/sup> \u2013 b2<\/sup>), (a \u2013 b), (a-b)\/(a+b), \u2026 to n terms<\/p>\n\n\n\n

We know that, sum of GP for n terms = a(rn<\/sup> \u2013 1)\/(r \u2013 1)<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

a = (a2<\/sup> \u2013 b2<\/sup>), r = t2<\/sub>\/t1<\/sub> = (a-b)\/(a2<\/sup> \u2013 b2<\/sup>) = (a-b)\/(a-b) (a+b) = 1\/(a+b), n = n<\/p>\n\n\n\n

Now let us substitute the values in<\/p>\n\n\n\n

a(rn<\/sup> \u2013 1)\/(r \u2013 1) =<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(v) <\/strong>4, 2, 1, \u00bd \u2026 to 10 terms<\/p>\n\n\n\n

We know that, sum of GP for n terms = a(rn<\/sup> \u2013 1)\/(r \u2013 1)<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

a = 4, r = t2<\/sub>\/t1<\/sub> = 2\/4 = 1\/2, n = 10<\/p>\n\n\n\n

Now let us substitute the values in<\/p>\n\n\n\n

a(rn<\/sup> \u2013 1)\/(r \u2013 1) = 4 ((1\/2)10<\/sup> \u2013 1)\/((1\/2)-1)<\/p>\n\n\n\n

= 4 ((1\/2)10<\/sup> \u2013 1)\/((1-2)\/2)<\/p>\n\n\n\n

= 4 ((1\/2)10<\/sup> \u2013 1)\/(-1\/2)<\/p>\n\n\n\n

= 4 ((1\/2)10<\/sup> \u2013 1) \u00d7 -2\/1<\/p>\n\n\n\n

= -8 [1\/1024 -1]<\/p>\n\n\n\n

= -8 [1 \u2013 1024]\/1024<\/p>\n\n\n\n

= -8 [-1023]\/1024<\/p>\n\n\n\n

= 1023\/128<\/p>\n\n\n\n

2. Find the sum of the following geometric series :
(i) 0.15 + 0.015 + 0.0015 + \u2026 to 8 terms;<\/strong><\/p>\n\n\n\n

(ii) \u221a2 + 1\/\u221a2 + 1\/2\u221a2 + \u2026. to 8 terms;<\/strong><\/p>\n\n\n\n

(iii) 2\/9 \u2013 1\/3 + \u00bd \u2013 \u00be + \u2026 to 5 terms;<\/strong><\/p>\n\n\n\n

(iv) (x + y) + (x2<\/sup> + xy + y2<\/sup>) + (x3<\/sup> + x2<\/sup> y + xy2<\/sup> + y3<\/sup>) + \u2026. to n terms ;<\/strong><\/p>\n\n\n\n

(v) 3\/5 + 4\/52<\/sup> + 3\/53<\/sup> + 4\/54<\/sup> + \u2026 to 2n terms;<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) <\/strong>0.15 + 0.015 + 0.0015 + \u2026 to 8 terms<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

a = 0.15<\/p>\n\n\n\n

r = t2<\/sub>\/t1<\/sub> = 0.015\/0.15 = 0.1 = 1\/10<\/p>\n\n\n\n

n = 8<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Sum of GP for n terms = a(1 \u2013 rn<\/sup> )\/(1 \u2013 r)<\/p>\n\n\n\n

a(1 \u2013 rn<\/sup> )\/(1 \u2013 r) = 0.15 (1 \u2013 (1\/10)8<\/sup>) \/ (1 \u2013 (1\/10))<\/p>\n\n\n\n

= 0.15 (1 \u2013 1\/108<\/sup>) \/ (1\/10)<\/p>\n\n\n\n

= 1\/6 (1 \u2013 1\/108<\/sup>)<\/p>\n\n\n\n

(ii) <\/strong>\u221a2 + 1\/\u221a2 + 1\/2\u221a2 + \u2026. to 8 terms;<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

a = \u221a2<\/p>\n\n\n\n

r = t2<\/sub>\/t1<\/sub> = (1\/\u221a2)\/\u221a2 = 1\/2<\/p>\n\n\n\n

n = 8<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Sum of GP for n terms = a(1 \u2013 rn<\/sup> )\/(1 \u2013 r)<\/p>\n\n\n\n

a(1 \u2013 rn<\/sup> )\/(1 \u2013 r) = \u221a2 (1 \u2013 (1\/2)8<\/sup>) \/ (1 \u2013 (1\/2))<\/p>\n\n\n\n

= \u221a2 (1 \u2013 1\/256) \/ (1\/2)<\/p>\n\n\n\n

= \u221a2 ((256 \u2013 1)\/256) \u00d7 2<\/p>\n\n\n\n

= \u221a2 (255\u00d72)\/256<\/p>\n\n\n\n

= (255\u221a2)\/128<\/p>\n\n\n\n

(iii) <\/strong>2\/9 \u2013 1\/3 + \u00bd \u2013 \u00be + \u2026 to 5 terms;<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

a = 2\/9<\/p>\n\n\n\n

r = t2<\/sub>\/t1<\/sub> = (-1\/3) \/ (2\/9) = -3\/2<\/p>\n\n\n\n

n = 5<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Sum of GP for n terms = a(1 \u2013 rn<\/sup> )\/(1 \u2013 r)<\/p>\n\n\n\n

a(1 \u2013 rn<\/sup> )\/(1 \u2013 r) = (2\/9) (1 \u2013 (-3\/2)5<\/sup>) \/ (1 \u2013 (-3\/2))<\/p>\n\n\n\n

= (2\/9) (1 + (3\/2)5<\/sup>) \/ (1 + 3\/2)<\/p>\n\n\n\n

= (2\/9) (1 + (3\/2)5<\/sup>) \/ (5\/2)<\/p>\n\n\n\n

= (2\/9) (1 + 243\/32) \/ (5\/2)<\/p>\n\n\n\n

= (2\/9) ((32+243)\/32) \/ (5\/2)<\/p>\n\n\n\n

= (2\/9) (275\/32) \u00d7 2\/5<\/p>\n\n\n\n

= 55\/72<\/p>\n\n\n\n

(iv) <\/strong>(x + y) + (x2<\/sup> + xy + y2<\/sup>) + (x3<\/sup> + x2<\/sup> y + xy2<\/sup> + y3<\/sup>) + \u2026. to n terms;<\/p>\n\n\n\n

Let Sn<\/sub> = (x + y) + (x2<\/sup> + xy + y2<\/sup>) + (x3<\/sup> + x2<\/sup> y + xy2<\/sup> + y3<\/sup>) + \u2026. to n terms<\/p>\n\n\n\n

Let us multiply and divide by (x \u2013 y) we get,<\/p>\n\n\n\n

Sn<\/sub> = 1\/(x \u2013 y) [(x + y) (x \u2013 y) + (x2<\/sup> + xy + y2<\/sup>) (x \u2013 y) \u2026 upto n terms]<\/p>\n\n\n\n

(x \u2013 y) Sn<\/sub> = (x2<\/sup> \u2013 y2<\/sup>) + x3<\/sup> + x2<\/sup>y + xy2<\/sup> \u2013 x2<\/sup>y \u2013 xy2<\/sup> \u2013 y3<\/sup>..upto n terms<\/p>\n\n\n\n

(x \u2013 y) Sn =<\/sub> (x2<\/sup> + x3<\/sup> + x4<\/sup>+\u2026n terms) \u2013 (y2<\/sup> + y3<\/sup> + y4<\/sup> +\u2026n terms)<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Sum of GP for n terms = a(1 \u2013 rn<\/sup> )\/(1 \u2013 r)<\/p>\n\n\n\n

We have two G.Ps in above sum, so,<\/p>\n\n\n\n

(x \u2013 y) Sn<\/sub> = x2<\/sup> [(xn<\/sup> \u2013 1)\/ (x \u2013 1)] \u2013 y2<\/sup> [(yn<\/sup> \u2013 1)\/ (y \u2013 1)]<\/p>\n\n\n\n

Sn<\/sub> = 1\/(x-y) {x2<\/sup> [(xn<\/sup> \u2013 1)\/ (x \u2013 1)] \u2013 y2<\/sup> [(yn<\/sup> \u2013 1)\/ (y \u2013 1)]}<\/p>\n\n\n\n

(v) <\/strong>3\/5 + 4\/52<\/sup> + 3\/53<\/sup> + 4\/54<\/sup> + \u2026 to 2n terms;<\/p>\n\n\n\n

The series can be written as:<\/p>\n\n\n\n

3 (1\/5 + 1\/53<\/sup> + 1\/55<\/sup>+ \u2026 to n terms) + 4 (1\/52<\/sup> + 1\/54<\/sup> + 1\/56<\/sup> + \u2026 to n terms)<\/p>\n\n\n\n

Firstly let us consider 3 (1\/5 + 1\/53<\/sup> + 1\/55<\/sup>+ \u2026 to n terms)<\/p>\n\n\n\n

So, a = 1\/5<\/p>\n\n\n\n

r = t2<\/sub>\/t1<\/sub> = 1\/52<\/sup> = 1\/25<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Sum of GP for n terms = a(1 \u2013 rn<\/sup> )\/(1 \u2013 r)<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Now, Let us consider 4 (1\/52<\/sup> + 1\/54<\/sup> + 1\/56<\/sup> + \u2026 to n terms)<\/p>\n\n\n\n

So, a = 1\/25<\/p>\n\n\n\n

r = t2<\/sub>\/t1<\/sub> = 1\/52<\/sup> = 1\/25<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Sum of GP for n terms = a(1 \u2013 rn<\/sup> )\/(1 \u2013 r)<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

3.<\/strong> Evaluate the following:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

= (2 + 31<\/sup>) + (2 + 32<\/sup>) + (2 + 33<\/sup>) + \u2026 + (2 + 311<\/sup>)<\/p>\n\n\n\n

= 2\u00d711 + 31<\/sup> + 32<\/sup> + 33<\/sup> + \u2026 + 311<\/sup><\/p>\n\n\n\n

= 22 + 3(311<\/sup> \u2013 1)\/(3 \u2013 1) [by using the formula, a(1 \u2013 rn<\/sup> )\/(1 \u2013 r)]<\/p>\n\n\n\n

= 22 + 3(311<\/sup> \u2013 1)\/2<\/p>\n\n\n\n

= [44 + 3(177147 \u2013 1)]\/2<\/p>\n\n\n\n

= [44 + 3(177146)]\/2<\/p>\n\n\n\n

= 265741<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

= (2 + 30<\/sup>) + (22<\/sup> + 3) + (23<\/sup> + 32<\/sup>) + \u2026 + (2n<\/sup> + 3n-1<\/sup>)<\/p>\n\n\n\n

= (2 + 22<\/sup> + 23<\/sup> + \u2026 + 2n<\/sup>) + (30<\/sup> + 31<\/sup> + 32<\/sup> + \u2026. + 3n-1<\/sup>)<\/p>\n\n\n\n

Firstly let us consider,<\/p>\n\n\n\n

(2 + 22<\/sup> + 23<\/sup> + \u2026 + 2n<\/sup>)<\/p>\n\n\n\n

Where, a = 2, r = 22<\/sup>\/2 = 4\/2 = 2, n = n<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Sum of GP for n terms = a(rn<\/sup> \u2013 1 )\/(r \u2013 1)<\/p>\n\n\n\n

= 2 (2n<\/sup> \u2013 1)\/(2 \u2013 1)<\/p>\n\n\n\n

= 2 (2n<\/sup> \u2013 1)<\/p>\n\n\n\n

Now, let us consider<\/p>\n\n\n\n

(30<\/sup> + 31<\/sup> + 32<\/sup> + \u2026. + 3n<\/sup>)<\/p>\n\n\n\n

Where, a = 30<\/sup> = 1, r = 3\/1 = 3, n = n<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Sum of GP for n terms = a(rn<\/sup> \u2013 1 )\/(r \u2013 1)<\/p>\n\n\n\n

= 1 (3n<\/sup> \u2013 1)\/ (3 \u2013 1)<\/p>\n\n\n\n

= (3n<\/sup> \u2013 1)\/2<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

= (2 + 22<\/sup> + 23<\/sup> + \u2026 + 2n<\/sup>) + (30<\/sup> + 31<\/sup> + 32<\/sup> + \u2026. + 3n<\/sup>)<\/p>\n\n\n\n

= 2 (2n<\/sup> \u2013 1) + (3n<\/sup> \u2013 1)\/2<\/p>\n\n\n\n

= \u00bd [2n+2<\/sup> + 3n<\/sup> \u2013 4 \u2013 1]<\/p>\n\n\n\n

= \u00bd [2n+2<\/sup> + 3n<\/sup> \u2013 5]<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

= 42<\/sup> + 43<\/sup> + 44<\/sup> + \u2026 + 410<\/sup><\/p>\n\n\n\n

Where, a = 42<\/sup> = 16, r = 43<\/sup>\/42<\/sup> = 4, n = 9<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Sum of GP for n terms = a(rn<\/sup> \u2013 1 )\/(r \u2013 1)<\/p>\n\n\n\n

= 16 (49<\/sup> \u2013 1)\/(4 \u2013 1)<\/p>\n\n\n\n

= 16 (49<\/sup> \u2013 1)\/3<\/p>\n\n\n\n

= 16\/3 [49<\/sup> \u2013 1]<\/p>\n\n\n\n

4. Find the sum of the following series :
(i) 5 + 55 + 555 + \u2026 to n terms.<\/strong><\/p>\n\n\n\n

(ii) 7 + 77 + 777 + \u2026 to n terms.<\/strong><\/p>\n\n\n\n

(iii) 9 + 99 + 999 + \u2026 to n terms.<\/strong><\/p>\n\n\n\n

(iv) 0.5 + 0.55 + 0.555 + \u2026. to n terms<\/strong><\/p>\n\n\n\n

(v) 0.6 + 0.66 + 0.666 + \u2026. to n terms.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) <\/strong>5 + 55 + 555 + \u2026 to n terms.<\/p>\n\n\n\n

Let us take 5 as a common term so we get,<\/p>\n\n\n\n

5 [1 + 11 + 111 + \u2026 n terms]<\/p>\n\n\n\n

Now multiply and divide by 9 we get,<\/p>\n\n\n\n

5\/9 [9 + 99 + 999 + \u2026 n terms]<\/p>\n\n\n\n

5\/9 [(10 \u2013 1) + (102<\/sup> \u2013 1) + (103<\/sup> \u2013 1) + \u2026 n terms]<\/p>\n\n\n\n

5\/9 [(10 + 102<\/sup> + 103<\/sup> + \u2026 n terms) \u2013 n]<\/p>\n\n\n\n

So the G.P is<\/p>\n\n\n\n

5\/9 [(10 + 102<\/sup> + 103<\/sup> + \u2026 n terms) \u2013 n]<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Sum of GP for n terms = a(rn<\/sup> \u2013 1 )\/(r \u2013 1)<\/p>\n\n\n\n

Where, a = 10, r = 102<\/sup>\/10 = 10, n = n<\/p>\n\n\n\n

a(rn<\/sup> \u2013 1 )\/(r \u2013 1) =<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(ii) <\/strong>7 + 77 + 777 + \u2026 to n terms.<\/p>\n\n\n\n

Let us take 7 as a common term so we get,<\/p>\n\n\n\n

7 [1 + 11 + 111 + \u2026 to n terms]<\/p>\n\n\n\n

Now multiply and divide by 9 we get,<\/p>\n\n\n\n

7\/9 [9 + 99 + 999 + \u2026 n terms]<\/p>\n\n\n\n

7\/9 [(10 \u2013 1) + (102<\/sup> \u2013 1) + (103<\/sup> \u2013 1) + \u2026 + (10n<\/sup> \u2013 1)]<\/p>\n\n\n\n

7\/9 [(10 + 102<\/sup> + 103<\/sup> + \u2026 +10n<\/sup>)] \u2013 7\/9 [(1 + 1 + 1 + \u2026 to n terms)]<\/p>\n\n\n\n

So the terms are in G.P<\/p>\n\n\n\n

Where, a = 10, r = 102<\/sup>\/10 = 10, n = n<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Sum of GP for n terms = a(rn<\/sup> \u2013 1 )\/(r \u2013 1)<\/p>\n\n\n\n

7\/9 [10 (10n<\/sup> \u2013 1)\/(10-1)] \u2013 n<\/p>\n\n\n\n

7\/9 [10\/9 (10n<\/sup> \u2013 1) \u2013 n]<\/p>\n\n\n\n

7\/81 [10 (10n<\/sup> \u2013 1) \u2013 n]<\/p>\n\n\n\n

7\/81 (10n+1<\/sup> \u2013 9n \u2013 10)<\/p>\n\n\n\n

(iii) <\/strong>9 + 99 + 999 + \u2026 to n terms.<\/p>\n\n\n\n

The given terms can be written as<\/p>\n\n\n\n

(10 \u2013 1) + (100 \u2013 1) + (1000 \u2013 1) + \u2026 + n terms<\/p>\n\n\n\n

(10 + 102<\/sup> + 103<\/sup> + \u2026 n terms) \u2013 n<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Sum of GP for n terms = a(rn<\/sup> \u2013 1 )\/(r \u2013 1)<\/p>\n\n\n\n

Where, a = 10, r = 10, n = n<\/p>\n\n\n\n

a(rn<\/sup> \u2013 1 )\/(r \u2013 1) = [10 (10n<\/sup> \u2013 1)\/(10-1)] \u2013 n<\/p>\n\n\n\n

= 10\/9 (10n<\/sup> \u2013 1) \u2013 n<\/p>\n\n\n\n

= 1\/9 [10n+1<\/sup> \u2013 10 \u2013 9n]<\/p>\n\n\n\n

= 1\/9 [10n+1<\/sup> \u2013 9n \u2013 10]<\/p>\n\n\n\n

(iv) <\/strong>0.5 + 0.55 + 0.555 + \u2026. to n terms<\/p>\n\n\n\n

Let us take 5 as a common term so we get,<\/p>\n\n\n\n

5(0.1 + 0.11 + 0.111 + \u2026n terms)<\/p>\n\n\n\n

Now multiply and divide by 9 we get,<\/p>\n\n\n\n

5\/9 [0.9 + 0.99 + 0.999 + \u2026+ to n terms]<\/p>\n\n\n\n

5\/9 [9\/10 + 9\/100 + 9\/1000 + \u2026 + n terms]<\/p>\n\n\n\n

This can be written as<\/p>\n\n\n\n

5\/9 [(1 \u2013 1\/10) + (1 \u2013 1\/100) + (1 \u2013 1\/1000) + \u2026 + n terms]<\/p>\n\n\n\n

5\/9 [n \u2013 {1\/10 + 1\/102<\/sup> + 1\/103<\/sup> + \u2026 + n terms}]<\/p>\n\n\n\n

5\/9 [n \u2013 1\/10 {1-(1\/10)n<\/sup>}\/{1 \u2013 1\/10}]<\/p>\n\n\n\n

5\/9 [n \u2013 1\/9 (1 \u2013 1\/10n<\/sup>)]<\/p>\n\n\n\n

(v) <\/strong>0.6 + 0.66 + 0.666 + \u2026. to n terms.<\/p>\n\n\n\n

Let us take 6 as a common term so we get,<\/p>\n\n\n\n

6(0.1 + 0.11 + 0.111 + \u2026n terms)<\/p>\n\n\n\n

Now multiply and divide by 9 we get,<\/p>\n\n\n\n

6\/9 [0.9 + 0.99 + 0.999 + \u2026+ n terms]<\/p>\n\n\n\n

6\/9 [9\/10 + 9\/100 + 9\/1000 + \u2026+ n terms]<\/p>\n\n\n\n

This can be written as<\/p>\n\n\n\n

6\/9 [(1 \u2013 1\/10) + (1 \u2013 1\/100) + (1 \u2013 1\/1000) + \u2026 + n terms]<\/p>\n\n\n\n

6\/9 [n \u2013 {1\/10 + 1\/102<\/sup> + 1\/103<\/sup> + \u2026 + n terms}]<\/p>\n\n\n\n

6\/9 [n \u2013 1\/10 {1-(1\/10)n<\/sup>}\/{1 \u2013 1\/10}]<\/p>\n\n\n\n

6\/9 [n \u2013 1\/9 (1 \u2013 1\/10n<\/sup>)]<\/p>\n\n\n\n

5. How many terms of the G.P. 3, 3\/2, \u00be, \u2026 Be taken together to make 3069\/512 ?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

Sum of G.P = 3069\/512<\/p>\n\n\n\n

Where, a = 3, r = (3\/2)\/3 = 1\/2, n = ?<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Sum of GP for n terms = a(rn<\/sup> \u2013 1 )\/(r \u2013 1)<\/p>\n\n\n\n

3069\/512 = 3 ((1\/2)n<\/sup> \u2013 1)\/ (1\/2 \u2013 1)<\/p>\n\n\n\n

3069\/512\u00d73\u00d72 = 1 \u2013 (1\/2)n<\/sup><\/p>\n\n\n\n

3069\/3072 \u2013 1 = \u2013 (1\/2)n<\/sup><\/p>\n\n\n\n

(3069 \u2013 3072)\/3072 = \u2013 (1\/2)n<\/sup><\/p>\n\n\n\n

-3\/3072 = \u2013 (1\/2)n<\/sup><\/p>\n\n\n\n

1\/1024 = (1\/2)n<\/sup><\/p>\n\n\n\n

(1\/2)10<\/sup> = (1\/2)n<\/sup><\/p>\n\n\n\n

10 = n<\/p>\n\n\n\n

\u2234 10 terms are required to make 3069\/512<\/p>\n\n\n\n

6. How many terms of the series 2 + 6 + 18 + \u2026. Must be taken to make the sum equal to 728?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

Sum of GP = 728<\/p>\n\n\n\n

Where, a = 2, r = 6\/2 = 3, n = ?<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Sum of GP for n terms = a(rn<\/sup> \u2013 1 )\/(r \u2013 1)<\/p>\n\n\n\n

728 = 2 (3n<\/sup> \u2013 1)\/(3-1)<\/p>\n\n\n\n

728 = 2 (3n<\/sup> \u2013 1)\/2<\/p>\n\n\n\n

728 = 3n<\/sup> \u2013 1<\/p>\n\n\n\n

729 = 3n<\/sup><\/p>\n\n\n\n

36<\/sup> = 3n<\/sup><\/p>\n\n\n\n

6 = n<\/p>\n\n\n\n

\u2234 6 terms are required to make a sum equal to 728<\/p>\n\n\n\n

7. How many terms of the sequence \u221a3, 3, 3\u221a3,\u2026 must be taken to make the sum 39+ 13\u221a3 ?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

Sum of GP = 39 + 13\u221a3<\/p>\n\n\n\n

Where, a =\u221a3, r = 3\/\u221a3 = \u221a3, n = ?<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Sum of GP for n terms = a(rn<\/sup> \u2013 1 )\/(r \u2013 1)<\/p>\n\n\n\n

39 + 13\u221a3 = \u221a3 (\u221a3n<\/sup> \u2013 1)\/ (\u221a3 \u2013 1)<\/p>\n\n\n\n

(39 + 13\u221a3) (\u221a3 \u2013 1) = \u221a3 (\u221a3n<\/sup> \u2013 1)<\/p>\n\n\n\n

Let us simplify we get,<\/p>\n\n\n\n

39\u221a3 \u2013 39 + 13(3) \u2013 13\u221a3 = \u221a3 (\u221a3n<\/sup> \u2013 1)<\/p>\n\n\n\n

39\u221a3 \u2013 39 + 39 \u2013 13\u221a3 = \u221a3 (\u221a3n<\/sup> \u2013 1)<\/p>\n\n\n\n

39\u221a3 \u2013 39 + 39 \u2013 13\u221a3 = \u221a3n+1<\/sup> \u2013 \u221a3<\/p>\n\n\n\n

26\u221a3 + \u221a3 = \u221a3n+1<\/sup><\/p>\n\n\n\n

27\u221a3 = \u221a3n+1<\/sup><\/p>\n\n\n\n

\u221a36<\/sup> \u221a3 = \u221a3n+1<\/sup><\/p>\n\n\n\n

6+1 = n + 1<\/p>\n\n\n\n

7 = n + 1<\/p>\n\n\n\n

7 \u2013 1 = n<\/p>\n\n\n\n

6 = n<\/p>\n\n\n\n

\u2234 6 terms are required to make a sum of 39 + 13\u221a3<\/p>\n\n\n\n

8. The sum of n terms of the G.P. 3, 6, 12, \u2026 is 381. Find the value of n.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

Sum of GP = 381<\/p>\n\n\n\n

Where, a = 3, r = 6\/3 = 2, n = ?<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Sum of GP for n terms = a(rn<\/sup> \u2013 1 )\/(r \u2013 1)<\/p>\n\n\n\n

381 = 3 (2n<\/sup> \u2013 1)\/ (2-1)<\/p>\n\n\n\n

381 = 3 (2n<\/sup> \u2013 1)<\/p>\n\n\n\n

381\/3 = 2n<\/sup> \u2013 1<\/p>\n\n\n\n

127 = 2n<\/sup> \u2013 1<\/p>\n\n\n\n

127 + 1 = 2n<\/sup><\/p>\n\n\n\n

128 = 2n<\/sup><\/p>\n\n\n\n

27<\/sup> = 2n<\/sup><\/p>\n\n\n\n

n = 7<\/p>\n\n\n\n

\u2234 value of n is 7<\/p>\n\n\n\n

9. The common ratio of a G.P. is 3, and the last term is 486. If the sum of these terms be 728, find the first term.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

Sum of GP = 728<\/p>\n\n\n\n

Where, r = 3, a = ?<\/p>\n\n\n\n

Firstly,<\/p>\n\n\n\n

Tn<\/sub> = arn-1<\/sup><\/p>\n\n\n\n

486 = a3n-1<\/sup><\/p>\n\n\n\n

486 = a3n<\/sup>\/3<\/p>\n\n\n\n

486 (3) = a3n<\/sup><\/p>\n\n\n\n

1458 = a3n<\/sup> \u2026. Equation (i)<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Sum of GP for n terms = a(rn<\/sup> \u2013 1 )\/(r \u2013 1)<\/p>\n\n\n\n

728 = a (3n<\/sup> \u2013 1)\/2<\/p>\n\n\n\n

1456 = a3n<\/sup> \u2013 a \u2026 equation (2)<\/p>\n\n\n\n

Subtracting equation (1) from (2) we get<\/p>\n\n\n\n

1458 \u2013 1456 = a.3n<\/sup> \u2013 a.3n<\/sup> + a<\/p>\n\n\n\n

a = 2.<\/p>\n\n\n\n

\u2234 The first term is 2<\/p>\n\n\n\n

10. The ratio of the sum of the first three terms is to that of the first 6 terms of a G.P. is 125 : 152. Find the common ratio.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

Sum of G.P of 3 terms is 125<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Sum of GP for n terms = a(rn<\/sup> \u2013 1 )\/(r \u2013 1)<\/p>\n\n\n\n

125 = a (rn<\/sup> \u2013 1)\/(r-1)<\/p>\n\n\n\n

125 = a (r3<\/sup> \u2013 1)\/ (r-1) \u2026 equation (1)<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

Sum of G.P of 6 terms is 152<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Sum of GP for n terms = a(rn<\/sup> \u2013 1 )\/(r \u2013 1)<\/p>\n\n\n\n

152 = a (rn<\/sup> \u2013 1)\/(r-1)<\/p>\n\n\n\n

152 = a (r6<\/sup> \u2013 1)\/ (r-1) \u2026 equation (2)<\/p>\n\n\n\n

Let us divide equation (i) by (ii) we get,<\/p>\n\n\n\n

125\/152 = [a (r3<\/sup> \u2013 1)\/ (r-1)] \/ [a (r6<\/sup> \u2013 1)\/ (r-1)]<\/p>\n\n\n\n

125\/152 = (r3<\/sup> \u2013 1)\/(r6<\/sup> \u2013 1)<\/p>\n\n\n\n

125\/152 = (r3<\/sup> \u2013 1)\/[(r3<\/sup> \u2013 1) (r3<\/sup> + 1)]<\/p>\n\n\n\n

125\/152 = 1\/(r3<\/sup> + 1)<\/p>\n\n\n\n

125(r3<\/sup> + 1) = 152<\/p>\n\n\n\n

125r3<\/sup> + 125 = 152<\/p>\n\n\n\n

125r3<\/sup> = 152 \u2013 125<\/p>\n\n\n\n

125r3<\/sup> = 27<\/p>\n\n\n\n

r3<\/sup> = 27\/125<\/p>\n\n\n\n

r3<\/sup> = 33<\/sup>\/53<\/sup><\/p>\n\n\n\n

r = 3\/5<\/p>\n\n\n\n

\u2234 The common ratio is 3\/5<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 20.4 PAGE NO: 20.39<\/p>\n\n\n\n

1. Find the sum of the following series to infinity:<\/strong><\/p>\n\n\n\n

(i) 1 \u2013 1\/3 + 1\/32<\/sup> \u2013 1\/33<\/sup> + 1\/34<\/sup> + \u2026 \u221e<\/strong><\/p>\n\n\n\n

(ii) 8 + 4\u221a2 + 4 + \u2026. \u221e<\/strong><\/p>\n\n\n\n

(iii) 2\/5 + 3\/52<\/sup> + 2\/53<\/sup> + 3\/54<\/sup> + \u2026. \u221e<\/strong><\/p>\n\n\n\n

(iv) 10 \u2013 9 + 8.1 \u2013 7.29 + \u2026. \u221e<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) <\/strong>1 \u2013 1\/3 + 1\/32<\/sup> \u2013 1\/33<\/sup> + 1\/34<\/sup> + \u2026 \u221e<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

S\u221e<\/sub> = 1 \u2013 1\/3 + 1\/32<\/sup> \u2013 1\/33<\/sup> + 1\/34<\/sup> + \u2026 \u221e<\/p>\n\n\n\n

Where, a = 1, r = -1\/3<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

S\u221e<\/sub> = a\/(1 \u2013 r)<\/p>\n\n\n\n

= 1 \/ (1 \u2013 (-1\/3))<\/p>\n\n\n\n

= 1\/ (1 + 1\/3)<\/p>\n\n\n\n

= 1\/ ((3+1)\/3)<\/p>\n\n\n\n

= 1\/ (4\/3)<\/p>\n\n\n\n

= \u00be<\/p>\n\n\n\n

(ii) <\/strong>8 + 4\u221a2 + 4 + \u2026. \u221e<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

S\u221e<\/sub> = 8 + 4\u221a2 + 4 + \u2026. \u221e<\/p>\n\n\n\n

Where, a = 8, r = 4\/4\u221a2 = 1\/\u221a2<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

S\u221e<\/sub> = a\/(1 \u2013 r)<\/p>\n\n\n\n

= 8 \/ (1 \u2013 (1\/\u221a2))<\/p>\n\n\n\n

= 8 \/ ((\u221a2 \u2013 1)\/\u221a2)<\/p>\n\n\n\n

= 8\u221a2 \/(\u221a2 \u2013 1)<\/p>\n\n\n\n

Multiply and divide with \u221a2 + 1 we get,<\/p>\n\n\n\n

= 8\u221a2 \/(\u221a2 \u2013 1) \u00d7 (\u221a2 + 1)\/( \u221a2 + 1)<\/p>\n\n\n\n

= 8 (2 + \u221a2)\/(2-1)<\/p>\n\n\n\n

= 8 (2 + \u221a2)<\/p>\n\n\n\n

(iii) <\/strong>2\/5 + 3\/52<\/sup> + 2\/53<\/sup> + 3\/54<\/sup> + \u2026. \u221e<\/p>\n\n\n\n

The given terms can be written as,<\/p>\n\n\n\n

(2\/5 + 2\/53<\/sup> + \u2026) + (3\/52<\/sup> + 3\/54<\/sup> + \u2026)<\/p>\n\n\n\n

(a = 2\/5, r = 1\/25) and (a = 3\/25, r = 1\/25)<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

S\u221e<\/sub> = a\/(1 \u2013 r)<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(iv) <\/strong>10 \u2013 9 + 8.1 \u2013 7.29 + \u2026. \u221e<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

S\u221e<\/sub> = 8 + 4\u221a2 + 4 + \u2026. \u221e<\/p>\n\n\n\n

Where, a = 10, r = -9\/10<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

S\u221e<\/sub> = a\/(1 \u2013 r)<\/p>\n\n\n\n

= 10 \/ (1 \u2013 (-9\/10))<\/p>\n\n\n\n

= 10 \/ (1 + 9\/10)<\/p>\n\n\n\n

= 10 \/ ((10+9)\/10)<\/p>\n\n\n\n

= 10 \/ (19\/10)<\/p>\n\n\n\n

= 100\/19<\/p>\n\n\n\n

= 5.263<\/p>\n\n\n\n

2. Prove that :
(91\/3<\/sup> . 91\/9<\/sup> . 91\/27<\/sup> \u2026.\u221e) = 3.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let us consider the LHS<\/p>\n\n\n\n

(91\/3<\/sup> . 91\/9<\/sup> . 91\/27<\/sup> \u2026.\u221e)<\/p>\n\n\n\n

This can be written as<\/p>\n\n\n\n

91\/3 + 1\/9 + 1\/27 + \u2026\u221e<\/sup><\/p>\n\n\n\n

So let us consider m = 1\/3 + 1\/9 + 1\/27 + \u2026 \u221e<\/p>\n\n\n\n

Where, a = 1\/3, r = (1\/9) \/ (1\/3) = 1\/3<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

S\u221e<\/sub> = a\/(1 \u2013 r)<\/p>\n\n\n\n

= (1\/3) \/ (1 \u2013 (1\/3))<\/p>\n\n\n\n

= (1\/3) \/ ((3-1)\/3)<\/p>\n\n\n\n

= (1\/3) \/ (2\/3)<\/p>\n\n\n\n

= \u00bd<\/p>\n\n\n\n

So, 9m<\/sup> = 91\/2<\/sup> = 3 = RHS<\/p>\n\n\n\n

Hence proved.<\/p>\n\n\n\n

3. Prove that :<\/strong><\/p>\n\n\n\n

(21\/4<\/sup> .41\/8<\/sup> . 81\/16<\/sup>. 161\/32<\/sup>\u2026.\u221e) = 2.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let us consider the LHS<\/p>\n\n\n\n

(21\/4<\/sup> .41\/8<\/sup> . 81\/16<\/sup>. 161\/32<\/sup>\u2026.\u221e)<\/p>\n\n\n\n

This can be written as<\/p>\n\n\n\n

21\/4<\/sup> . 22\/8<\/sup> . 23\/16<\/sup> . 21\/8<\/sup> \u2026 \u221e<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

21\/4 + 2\/8 + 3\/16 + 1\/8 + \u2026\u221e<\/sup><\/p>\n\n\n\n

So let us consider 2x<\/sup>, where x = \u00bc + 2\/8 + 3\/16 + 1\/8 + \u2026 \u221e \u2026. (1)<\/p>\n\n\n\n

Multiply both sides of the equation with 1\/2, we get<\/p>\n\n\n\n

x\/2 = \u00bd (\u00bc + 2\/8 + 3\/16 + 1\/8 + \u2026 \u221e)<\/p>\n\n\n\n

= 1\/8 + 2\/16 + 3\/32 + \u2026 + \u221e \u2026. (2)<\/p>\n\n\n\n

Now, subtract (2) from (1) we get,<\/p>\n\n\n\n

x \u2013 x\/2 = (\u00bc + 2\/8 + 3\/16 + 1\/8 + \u2026 \u221e) \u2013 (1\/8 + 2\/16 + 3\/32 + \u2026 + \u221e)<\/p>\n\n\n\n

By grouping similar terms,<\/p>\n\n\n\n

x\/2 = \u00bc + (2\/8 \u2013 1\/8) + (3\/16 \u2013 2\/16) + \u2026 \u221e<\/p>\n\n\n\n

x\/2 = \u00bc + 1\/8 + 1\/16 + \u2026 \u221e<\/p>\n\n\n\n

x = \u00bd + \u00bc + 1\/8 + 1\/16 + \u2026 \u221e<\/p>\n\n\n\n

Where, a = 1\/2, r = (1\/4) \/ (1\/2) = 1\/2<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

S\u221e<\/sub> = a\/(1 \u2013 r)<\/p>\n\n\n\n

= (1\/2) \/ (1 \u2013 1\/2)<\/p>\n\n\n\n

= (1\/2) \/ ((2-1)\/2)<\/p>\n\n\n\n

= (1\/2) \/ (1\/2)<\/p>\n\n\n\n

= 1<\/p>\n\n\n\n

From equation (1), 2x<\/sup> = 21<\/sup> = 2 = RHS<\/p>\n\n\n\n

Hence proved.<\/p>\n\n\n\n

4. If Sp<\/sub> denotes the sum of the series 1 + rp<\/sup> + r2p<\/sup> + \u2026 to \u221e and sp<\/sub> the sum of the series 1 \u2013 rp<\/sup> + r2p<\/sup> \u2013 \u2026 to \u221e, prove that sp<\/sub> + Sp<\/sub> = 2 S2p<\/sub>.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

Sp<\/sub> = 1 + rp<\/sup> + r2p<\/sup> + \u2026 \u221e<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

S\u221e<\/sub> = a\/(1 \u2013 r)<\/p>\n\n\n\n

Where, a = 1, r = rp<\/sup><\/p>\n\n\n\n

So,<\/p>\n\n\n\n

Sp<\/sub> = 1 \/ (1 \u2013 rp<\/sup>)<\/p>\n\n\n\n

Similarly, sp<\/sub> = 1 \u2013 rp<\/sup> + r2p<\/sup> \u2013 \u2026 \u221e<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

S\u221e<\/sub> = a\/(1 \u2013 r)<\/p>\n\n\n\n

Where, a = 1, r = -rp<\/sup><\/p>\n\n\n\n

So,<\/p>\n\n\n\n

Sp<\/sub> = 1 \/ (1 \u2013 (-rp<\/sup>))<\/p>\n\n\n\n

= 1 \/ (1 + rp<\/sup>)<\/p>\n\n\n\n

Now, Sp<\/sub> + sp<\/sub> = [1 \/ (1 \u2013 rp<\/sup>)] + [1 \/ (1 + rp<\/sup>)]<\/p>\n\n\n\n

2S2p<\/sub> = [(1 \u2013 rp<\/sup>) + (1 + rp<\/sup>)] \/ (1 \u2013 r2p<\/sup>)<\/p>\n\n\n\n

= 2 \/(1 \u2013 r2p<\/sup>)<\/p>\n\n\n\n

\u2234 2S2p<\/sub> = Sp<\/sub> + Sp<\/sub><\/p>\n\n\n\n

5. Find the sum of the terms of an infinite decreasing G.P. in which all the terms are positive, the first term is 4, and the difference between the third and fifth term is equal to 32\/81.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let \u2018a\u2019 be the first term of GP and \u2018r\u2019 be the common ratio.<\/p>\n\n\n\n

We know that nth term of a GP is given by-<\/p>\n\n\n\n

an<\/sub> = arn-1<\/sup><\/p>\n\n\n\n

As, a = 4 (given)<\/p>\n\n\n\n

And a5<\/sub> \u2013 a3<\/sub> = 32\/81 (given)<\/p>\n\n\n\n

4r4<\/sup> \u2013 4r2<\/sup> = 32\/81<\/p>\n\n\n\n

4r2<\/sup>(r2<\/sup> \u2013 1) = 32\/81<\/p>\n\n\n\n

r2<\/sup>(r2<\/sup> \u2013 1) = 8\/81<\/p>\n\n\n\n

Let us denote r2<\/sup> with y<\/p>\n\n\n\n

81y(y-1) = 8<\/p>\n\n\n\n

81y2<\/sup> \u2013 81y \u2013 8 = 0<\/p>\n\n\n\n

Using the formula of the quadratic equation to solve the equation, we get<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

y = 18\/162 = 1\/9 or<\/p>\n\n\n\n

y = 144\/162<\/p>\n\n\n\n

= 8\/9<\/p>\n\n\n\n

So, r2<\/sup> = 1\/9 or 8\/9<\/p>\n\n\n\n

= 1\/3 or 2\u221a<\/strong>2\/3<\/p>\n\n\n\n

We know that,<\/p>\n\n\n\n

Sum of infinite, S\u221e<\/sub> = a\/(1 \u2013 r)<\/p>\n\n\n\n

Where, a = 4, r = 1\/3<\/p>\n\n\n\n

S\u221e<\/sub> = 4 \/ (1 \u2013 (1\/3))<\/p>\n\n\n\n

= 4 \/ ((3-1)\/3)<\/p>\n\n\n\n

= 4 \/ (2\/3)<\/p>\n\n\n\n

= 12\/2<\/p>\n\n\n\n

= 6<\/p>\n\n\n\n

Sum of infinite, S\u221e<\/sub> = a\/(1 \u2013 r)<\/p>\n\n\n\n

Where, a = 4, r = 2\u221a<\/strong>2\/3<\/p>\n\n\n\n

S\u221e<\/sub> = 4 \/ (1 \u2013 (2\u221a<\/strong>2\/3))<\/p>\n\n\n\n

= 12 \/ (3 \u2013 2\u221a<\/strong>2)<\/p>\n\n\n\n

6. Express the recurring decimal 0.125125125 \u2026 as a rational number.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

0.125125125<\/p>\n\n\n\n

So, 0.125125125 =
<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

= 0.125 + 0.000125 + 0.000000125 + \u2026<\/p>\n\n\n\n

This can be written as<\/p>\n\n\n\n

125\/103<\/sup> + 125\/106<\/sup> + 125\/109<\/sup> + \u2026<\/p>\n\n\n\n

125\/103<\/sup> [1 + 1\/103<\/sup> + 1\/106<\/sup> + \u2026]<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

S\u221e<\/sub> = a\/(1 \u2013 r)<\/p>\n\n\n\n

125\/103<\/sup> [1 \/ (1 \u2013 1\/1000)]<\/p>\n\n\n\n

125\/103<\/sup> [1 \/ ((1000 \u2013 1)\/1000))]<\/p>\n\n\n\n

125\/103<\/sup> [1 \/ (999\/1000)]<\/p>\n\n\n\n

125\/1000 (1000\/999)<\/p>\n\n\n\n

125\/999<\/p>\n\n\n\n

\u2234 The decimal 0.125125125 can be expressed in rational number as 125\/999<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 20.5 PAGE NO: 20.45<\/p>\n\n\n\n

1. If a, b, c are in G.P., prove that log a, log b, log c are in A.P.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that a, b and c are in G.P.<\/p>\n\n\n\n

b2<\/sup> = ac {using property of geometric mean}<\/p>\n\n\n\n

(b2<\/sup>)n<\/sup> = (ac)n<\/sup><\/p>\n\n\n\n

b2n<\/sup> = an<\/sup> cn<\/sup><\/p>\n\n\n\n

Now, apply log on both the sides we get,<\/p>\n\n\n\n

log b2n<\/sup> = log (an<\/sup> cn<\/sup>)<\/p>\n\n\n\n

log (bn<\/sup>)2<\/sup> = log an<\/sup> + log cn<\/sup><\/p>\n\n\n\n

2 log bn<\/sup> = log an<\/sup> + log cn<\/sup><\/p>\n\n\n\n

\u2234 log an<\/sup>, log bn<\/sup>, log cn<\/sup> are in A.P<\/p>\n\n\n\n

2. If a, b, c are in G.P., prove that 1\/loga<\/sub> m , 1\/logb<\/sub> m, 1\/logc<\/sub> m are in A.P.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

a, b and c are in GP<\/p>\n\n\n\n

b2<\/sup> = ac {property of geometric mean}<\/p>\n\n\n\n

Apply log on both sides with base m<\/p>\n\n\n\n

logm<\/sub> b2<\/sup> = logm<\/sub> ac<\/p>\n\n\n\n

logm<\/sub> b2<\/sup> = logm<\/sub> a + logm<\/sub> c {using property of log}<\/p>\n\n\n\n

2logm<\/sub> b = logm<\/sub> a + logm<\/sub> c<\/p>\n\n\n\n

2\/logb<\/sub> m = 1\/loga<\/sub> m + 1\/logc<\/sub> m<\/p>\n\n\n\n

\u2234 1\/loga<\/sub> m , 1\/logb<\/sub> m, 1\/logc<\/sub> m are in A.P.<\/p>\n\n\n\n

3. Find k such that k + 9, k \u2013 6 and 4 form three consecutive terms of a G.P.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let a = k + 9; b = k\u22126; and c = 4;<\/p>\n\n\n\n

We know that a, b and c are in GP, then<\/p>\n\n\n\n

b2<\/sup> = ac {using property of geometric mean}<\/p>\n\n\n\n

(k \u2212 6)2<\/sup> = 4(k + 9)<\/p>\n\n\n\n

k2<\/sup> \u2013 12k + 36 = 4k + 36<\/p>\n\n\n\n

k2<\/sup> \u2013 16k = 0<\/p>\n\n\n\n

k = 0 or k = 16<\/p>\n\n\n\n

4. Three numbers are in A.P., and their sum is 15. If 1, 3, 9 be added to them respectively, they from a G.P. find the numbers.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let the first term of an A.P. be \u2018a\u2019 and its common difference be\u2018d\u2019.<\/p>\n\n\n\n

a1<\/sub> + a2<\/sub> + a3<\/sub> = 15<\/p>\n\n\n\n

Where, the three number are: a, a + d, and a + 2d<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

a + a + d + a + 2d = 15<\/p>\n\n\n\n

3a + 3d = 15 or a + d = 5<\/p>\n\n\n\n

d = 5 \u2013 a \u2026 (i)<\/p>\n\n\n\n

Now, according to the question:<\/p>\n\n\n\n

a + 1, a + d + 3, and a + 2d + 9<\/p>\n\n\n\n

they are in GP, that is:<\/p>\n\n\n\n

(a+d+3)\/(a+1) = (a+2d+9)\/(a+d+3)<\/p>\n\n\n\n

(a + d + 3)2<\/sup> = (a + 2d + 9) (a + 1)<\/p>\n\n\n\n

a2<\/sup> + d2<\/sup> + 9 + 2ad + 6d + 6a = a2<\/sup> + a + 2da + 2d + 9a + 9<\/p>\n\n\n\n

(5 \u2013 a)2<\/sup> \u2013 4a + 4(5 \u2013 a) = 0<\/p>\n\n\n\n

25 + a2<\/sup> \u2013 10a \u2013 4a + 20 \u2013 4a = 0<\/p>\n\n\n\n

a2<\/sup> \u2013 18a + 45 = 0<\/p>\n\n\n\n

a2<\/sup> \u2013 15a \u2013 3a + 45 = 0<\/p>\n\n\n\n

a(a \u2013 15) \u2013 3(a \u2013 15) = 0<\/p>\n\n\n\n

a = 3 or a = 15<\/p>\n\n\n\n

d = 5 \u2013 a<\/p>\n\n\n\n

d = 5 \u2013 3 or d = 5 \u2013 15<\/p>\n\n\n\n

d = 2 or \u2013 10<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

For a = 3 and d = 2, the A.P is 3, 5, 7<\/p>\n\n\n\n

For a = 15 and d = -10, the A.P is 15, 5, -5<\/p>\n\n\n\n

\u2234 The numbers are 3, 5, 7 or 15, 5, \u2013 5<\/p>\n\n\n\n

5. The sum of three numbers which are consecutive terms of an A.P. is 21. If the second number is reduced by 1 and the third is increased by 1, we obtain three consecutive terms of a G.P. Find the numbers.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let the first term of an A.P. be \u2018a\u2019 and its common difference be\u2018d\u2019.<\/p>\n\n\n\n

a1<\/sub> + a2<\/sub> + a3<\/sub> = 21<\/p>\n\n\n\n

Where, the three number are: a, a + d, and a + 2d<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

3a + 3d = 21 or<\/p>\n\n\n\n

a + d = 7.<\/p>\n\n\n\n

d = 7 \u2013 a \u2026. (i)<\/p>\n\n\n\n

Now, according to the question:<\/p>\n\n\n\n

a, a + d \u2013 1, and a + 2d + 1<\/p>\n\n\n\n

they are now in GP, that is:<\/p>\n\n\n\n

(a+d-1)\/a = (a+2d+1)\/(a+d-1)<\/p>\n\n\n\n

(a + d \u2013 1)2<\/sup> = a(a + 2d + 1)<\/p>\n\n\n\n

a2<\/sup> + d2<\/sup> + 1 + 2ad \u2013 2d \u2013 2a = a2<\/sup> + a + 2da<\/p>\n\n\n\n

(7 \u2013 a)2<\/sup> \u2013 3a + 1 \u2013 2(7 \u2013 a) = 0<\/p>\n\n\n\n

49 + a2<\/sup> \u2013 14a \u2013 3a + 1 \u2013 14 + 2a = 0<\/p>\n\n\n\n

a2<\/sup> \u2013 15a + 36 = 0<\/p>\n\n\n\n

a2<\/sup> \u2013 12a \u2013 3a + 36 = 0<\/p>\n\n\n\n

a(a \u2013 12) \u2013 3(a \u2013 12) = 0<\/p>\n\n\n\n

a = 3 or a = 12<\/p>\n\n\n\n

d = 7 \u2013 a<\/p>\n\n\n\n

d = 7 \u2013 3 or d = 7 \u2013 12<\/p>\n\n\n\n

d = 4 or \u2013 5<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

For a = 3 and d = 4, the A.P is 3, 7, 11<\/p>\n\n\n\n

For a = 12 and d = -5, the A.P is 12, 7, 2<\/p>\n\n\n\n

\u2234 The numbers are 3, 7, 11 or 12, 7, 2<\/p>\n\n\n\n

6. The sum of three numbers a, b, c in A.P. is 18. If a and b are each increased by 4 and c is increased by 36, the new numbers form a G.P. Find a, b, c.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let the first term of an A.P. be \u2018a\u2019 and its common difference be\u2018d\u2019.<\/p>\n\n\n\n

b = a + d; c = a + 2d.<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

a + b + c = 18<\/p>\n\n\n\n

3a + 3d = 18 or a + d = 6.<\/p>\n\n\n\n

d = 6 \u2013 a \u2026 (i)<\/p>\n\n\n\n

Now, according to the question:<\/p>\n\n\n\n

a + 4, a + d + 4, and a + 2d + 36<\/p>\n\n\n\n

they are now in GP, that is:<\/p>\n\n\n\n

(a+d+4)\/(a+4) = (a+2d+36)\/(a+d+4)<\/p>\n\n\n\n

(a + d + 4)2<\/sup> = (a + 2d + 36)(a + 4)<\/p>\n\n\n\n

a2<\/sup> + d2<\/sup> + 16 + 8a + 2ad + 8d = a2<\/sup> + 4a + 2da + 36a + 144 + 8d<\/p>\n\n\n\n

d2<\/sup> \u2013 32a \u2013 128<\/p>\n\n\n\n

(6 \u2013 a)2<\/sup> \u2013 32a \u2013 128 = 0<\/p>\n\n\n\n

36 + a2<\/sup> \u2013 12a \u2013 32a \u2013 128 = 0<\/p>\n\n\n\n

a2<\/sup> \u2013 44a \u2013 92 = 0<\/p>\n\n\n\n

a2<\/sup> \u2013 46a + 2a \u2013 92 = 0<\/p>\n\n\n\n

a(a \u2013 46) + 2(a \u2013 46) = 0<\/p>\n\n\n\n

a = \u2013 2 or a = 46<\/p>\n\n\n\n

d = 6 \u2013a<\/p>\n\n\n\n

d = 6 \u2013 (\u2013 2) or d = 6 \u2013 46<\/p>\n\n\n\n

d = 8 or \u2013 40<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

For a = -2 and d = 8, the A.P is -2, 6, 14<\/p>\n\n\n\n

For a = 46 and d = -40, the A.P is 46, 6, -34<\/p>\n\n\n\n

\u2234 The numbers are \u2013 2, 6, 14 or 46, 6, \u2013 34<\/p>\n\n\n\n

7. The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an A.P. Find the numbers.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let the three numbers be a, ar, ar2<\/sup><\/p>\n\n\n\n

According to the question<\/p>\n\n\n\n

a + ar + ar2<\/sup> = 56 \u2026 (1)<\/p>\n\n\n\n

Let us subtract 1,7,21 we get,<\/p>\n\n\n\n

(a \u2013 1), (ar \u2013 7), (ar2<\/sup> \u2013 21)<\/p>\n\n\n\n

The above numbers are in AP.<\/p>\n\n\n\n

If three numbers are in AP, by the idea of the arithmetic mean, we can write 2b = a + c<\/p>\n\n\n\n

2 (ar \u2013 7) = a \u2013 1 + ar2<\/sup> \u2013 21<\/p>\n\n\n\n

= (ar2<\/sup> + a) \u2013 22<\/p>\n\n\n\n

2ar \u2013 14 = (56 \u2013 ar) \u2013 22<\/p>\n\n\n\n

2ar \u2013 14 = 34 \u2013 ar<\/p>\n\n\n\n

3ar = 48<\/p>\n\n\n\n

ar = 48\/3<\/p>\n\n\n\n

ar = 16<\/p>\n\n\n\n

a = 16\/r \u2026. (2)<\/p>\n\n\n\n

Now, substitute the value of a in equation (1) we get,<\/p>\n\n\n\n

(16 + 16r + 16r2<\/sup>)\/r = 56<\/p>\n\n\n\n

16 + 16r + 16r2<\/sup> = 56r<\/p>\n\n\n\n

16r2<\/sup> \u2013 40r + 16 = 0<\/p>\n\n\n\n

2r2<\/sup> \u2013 5r + 2 = 0<\/p>\n\n\n\n

2r2<\/sup> \u2013 4r \u2013 r + 2 = 0<\/p>\n\n\n\n

2r(r \u2013 2) \u2013 1(r \u2013 2) = 0<\/p>\n\n\n\n

(r \u2013 2) (2r \u2013 1) = 0<\/p>\n\n\n\n

r = 2 or 1\/2<\/p>\n\n\n\n

Substitute the value of r in equation (2) we get,<\/p>\n\n\n\n

a = 16\/r<\/p>\n\n\n\n

= 16\/2 or 16\/(1\/2)<\/p>\n\n\n\n

= 8 or 32<\/p>\n\n\n\n

\u2234 The three numbers are (a, ar, ar2<\/sup>) is (8, 16, 32)<\/p>\n\n\n\n

8. if a, b, c are in G.P., prove that:<\/strong><\/p>\n\n\n\n

(i) a(b2<\/sup> + c2<\/sup>) = c(a2<\/sup> + b2<\/sup>)<\/strong><\/p>\n\n\n\n

(ii) a2<\/sup>b2<\/sup>c2<\/sup> [1\/a3<\/sup> + 1\/b3<\/sup> + 1\/c3<\/sup>] = a3<\/sup> + b3<\/sup> + c3<\/sup><\/strong><\/p>\n\n\n\n

(iii) (a+b+c)2<\/sup> \/ (a2<\/sup> + b2<\/sup> + c2<\/sup>) = (a+b+c) \/ (a-b+c)<\/strong><\/p>\n\n\n\n

(iv) 1\/(a2<\/sup> \u2013 b2<\/sup>) + 1\/b2<\/sup> = 1\/(b2<\/sup> \u2013 c2<\/sup>)<\/strong><\/p>\n\n\n\n

(v) (a + 2b + 2c) (a \u2013 2b + 2c) = a2<\/sup> + 4c2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) <\/strong>a(b2<\/sup> + c2<\/sup>) = c(a2<\/sup> + b2<\/sup>)<\/p>\n\n\n\n

Given that a, b, c are in GP.<\/p>\n\n\n\n

By using the property of geometric mean,<\/p>\n\n\n\n

b2<\/sup> = ac<\/p>\n\n\n\n

Let us consider LHS: a(b2<\/sup> + c2<\/sup>)<\/p>\n\n\n\n

Now, substituting b2<\/sup> = ac, we get<\/p>\n\n\n\n

a(ac + c2<\/sup>)<\/p>\n\n\n\n

a2<\/sup>c + ac2<\/sup><\/p>\n\n\n\n

c(a2<\/sup> + ac)<\/p>\n\n\n\n

Substitute ac = b2<\/sup> we get,<\/p>\n\n\n\n

c(a2<\/sup> + b2<\/sup>) = RHS<\/p>\n\n\n\n

\u2234 LHS = RHS<\/p>\n\n\n\n

Hence proved.<\/p>\n\n\n\n

(ii) <\/strong>a2<\/sup>b2<\/sup>c2<\/sup> [1\/a3<\/sup> + 1\/b3<\/sup> + 1\/c3<\/sup>] = a3<\/sup> + b3<\/sup> + c3<\/sup><\/p>\n\n\n\n

Given that a, b, c are in GP.<\/p>\n\n\n\n

By using the property of geometric mean,<\/p>\n\n\n\n

b2<\/sup> = ac<\/p>\n\n\n\n

Let us consider LHS: a2<\/sup>b2<\/sup>c2<\/sup> [1\/a3<\/sup> + 1\/b3<\/sup> + 1\/c3<\/sup>]<\/p>\n\n\n\n

a2<\/sup>b2<\/sup>c2<\/sup>\/a3<\/sup> + a2<\/sup>b2<\/sup>c2<\/sup>\/b3<\/sup> + a2<\/sup>b2<\/sup>c2<\/sup>\/c3<\/sup><\/p>\n\n\n\n

b2<\/sup>c2<\/sup>\/a + a2<\/sup>c2<\/sup>\/b + a2<\/sup>b2<\/sup>\/c<\/p>\n\n\n\n

(ac)c2<\/sup>\/a + (b2<\/sup>)2<\/sup>\/b + a2<\/sup>(ac)\/c [by substituting the b2<\/sup> = ac]<\/p>\n\n\n\n

ac3<\/sup>\/a + b4<\/sup>\/b + a3<\/sup>c\/c<\/p>\n\n\n\n

c3<\/sup> + b3<\/sup> + a3<\/sup> = RHS<\/p>\n\n\n\n

\u2234 LHS = RHS<\/p>\n\n\n\n

Hence proved.<\/p>\n\n\n\n

(iii) <\/strong>(a+b+c)2<\/sup> \/ (a2<\/sup> + b2<\/sup> + c2<\/sup>) = (a+b+c) \/ (a-b+c)<\/p>\n\n\n\n

Given that a, b, c are in GP.<\/p>\n\n\n\n

By using the property of geometric mean,<\/p>\n\n\n\n

b2<\/sup> = ac<\/p>\n\n\n\n

Let us consider LHS: (a+b+c)2<\/sup> \/ (a2<\/sup> + b2<\/sup> + c2<\/sup>)<\/p>\n\n\n\n

(a+b+c)2<\/sup> \/ (a2<\/sup> + b2<\/sup> + c2<\/sup>) = (a+b+c)2<\/sup> \/ (a2<\/sup> \u2013 b2<\/sup> + c2<\/sup> + 2b2<\/sup>)<\/p>\n\n\n\n

= (a+b+c)2<\/sup> \/ (a2<\/sup> \u2013 b2<\/sup> + c2<\/sup> + 2ac) [Since, b2<\/sup> = ac]<\/p>\n\n\n\n

= (a+b+c)2<\/sup> \/ (a+b+c)(a-b+c) [Since, (a+b+c)(a-b+c) = a2<\/sup> \u2013 b2<\/sup> + c2<\/sup> + 2ac]<\/p>\n\n\n\n

= (a+b+c) \/ (a-b+c)<\/p>\n\n\n\n

= RHS<\/p>\n\n\n\n

\u2234 LHS = RHS<\/p>\n\n\n\n

Hence proved.<\/p>\n\n\n\n

(iv) <\/strong>1\/(a2<\/sup> \u2013 b2<\/sup>) + 1\/b2<\/sup> = 1\/(b2<\/sup> \u2013 c2<\/sup>)<\/p>\n\n\n\n

Given that a, b, c are in GP.<\/p>\n\n\n\n

By using the property of geometric mean,<\/p>\n\n\n\n

b2<\/sup> = ac<\/p>\n\n\n\n

Let us consider LHS: 1\/(a2<\/sup> \u2013 b2<\/sup>) + 1\/b2<\/sup><\/p>\n\n\n\n

Let us take LCM<\/p>\n\n\n\n

1\/(a2<\/sup> \u2013 b2<\/sup>) + 1\/b2<\/sup> = (b2<\/sup> + a2<\/sup> \u2013 b2<\/sup>)\/(a2<\/sup> \u2013 b2<\/sup>)b2<\/sup><\/p>\n\n\n\n

= a2 <\/sup>\/ (a2<\/sup>b2<\/sup> \u2013 b4<\/sup>)<\/p>\n\n\n\n

= a2<\/sup> \/ (a2<\/sup>b2<\/sup> \u2013 (b2<\/sup>)2<\/sup>)<\/p>\n\n\n\n

= a2<\/sup> \/ (a2<\/sup>b2<\/sup> \u2013 (ac)2<\/sup>) [Since, b2<\/sup> = ac]<\/p>\n\n\n\n

= a2<\/sup> \/ (a2<\/sup>b2<\/sup> \u2013 a2<\/sup>c2<\/sup>)<\/p>\n\n\n\n

= a2<\/sup> \/ a2<\/sup>(b2<\/sup> \u2013 c2<\/sup>)<\/p>\n\n\n\n

= 1\/ (b2<\/sup> \u2013 c2<\/sup>)<\/p>\n\n\n\n

= RHS<\/p>\n\n\n\n

\u2234 LHS = RHS<\/p>\n\n\n\n

Hence proved.<\/p>\n\n\n\n

(v) <\/strong>(a + 2b + 2c) (a \u2013 2b + 2c) = a2<\/sup> + 4c2<\/sup><\/p>\n\n\n\n

Given that a, b, c are in GP.<\/p>\n\n\n\n

By using the property of geometric mean,<\/p>\n\n\n\n

b2<\/sup> = ac<\/p>\n\n\n\n

Let us consider LHS: (a + 2b + 2c) (a \u2013 2b + 2c)<\/p>\n\n\n\n

Upon expansion we get,<\/p>\n\n\n\n

(a + 2b + 2c) (a \u2013 2b + 2c) = a2<\/sup> \u2013 2ab + 2ac + 2ab \u2013 4b2<\/sup> + 4bc + 2ac \u2013 4bc + 4c2<\/sup><\/p>\n\n\n\n

= a2<\/sup> + 4ac \u2013 4b2<\/sup> + 4c2<\/sup><\/p>\n\n\n\n

= a2<\/sup> + 4ac \u2013 4(ac) + 4c2<\/sup> [Since, b2<\/sup> = ac]<\/p>\n\n\n\n

= a2<\/sup> + 4c2<\/sup><\/p>\n\n\n\n

= RHS<\/p>\n\n\n\n

\u2234 LHS = RHS<\/p>\n\n\n\n

Hence proved.<\/p>\n\n\n\n

9. If a, b, c, d are in G.P., prove that:<\/strong><\/p>\n\n\n\n

(i) (ab \u2013 cd) \/ (b2<\/sup> \u2013 c2<\/sup>) = (a + c) \/ b<\/strong><\/p>\n\n\n\n

(ii) (a + b + c + d)2<\/sup> = (a + b)2<\/sup> + 2(b + c)2<\/sup> + (c + d)2<\/sup><\/strong><\/p>\n\n\n\n

(iii) (b + c) (b + d) = (c + a) (c + d)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) <\/strong>(ab \u2013 cd) \/ (b2<\/sup> \u2013 c2<\/sup>) = (a + c) \/ b<\/p>\n\n\n\n

Given that a, b, c are in GP.<\/p>\n\n\n\n

By using the property of geometric mean,<\/p>\n\n\n\n

b2<\/sup> = ac<\/p>\n\n\n\n

bc = ad<\/p>\n\n\n\n

c2<\/sup> = bd<\/p>\n\n\n\n

Let us consider LHS: (ab \u2013 cd) \/ (b2<\/sup> \u2013 c2<\/sup>)<\/p>\n\n\n\n

(ab \u2013 cd) \/ (b2<\/sup> \u2013 c2<\/sup>) = (ab \u2013 cd) \/ (ac \u2013 bd)<\/p>\n\n\n\n

= (ab \u2013 cd)b \/ (ac \u2013 bd)b<\/p>\n\n\n\n

= (ab2<\/sup> \u2013 bcd) \/ (ac \u2013 bd)b<\/p>\n\n\n\n

= [a(ac) \u2013 c(c2<\/sup>)] \/ (ac \u2013 bd)b<\/p>\n\n\n\n

= (a2<\/sup>c \u2013 c3<\/sup>) \/ (ac \u2013 bd)b<\/p>\n\n\n\n

= [c(a2<\/sup> \u2013 c2<\/sup>)] \/ (ac \u2013 bd)b<\/p>\n\n\n\n

= [(a+c) (ac \u2013 c2<\/sup>)] \/ (ac \u2013 bd)b<\/p>\n\n\n\n

= [(a+c) (ac \u2013 bd)] \/ (ac \u2013 bd)b<\/p>\n\n\n\n

= (a+c) \/ b<\/p>\n\n\n\n

= RHS<\/p>\n\n\n\n

\u2234 LHS = RHS<\/p>\n\n\n\n

Hence proved.<\/p>\n\n\n\n

(ii) <\/strong>(a + b + c + d)2<\/sup> = (a + b)2<\/sup> + 2(b + c)2<\/sup> + (c + d)2<\/sup><\/p>\n\n\n\n

Given that a, b, c are in GP.<\/p>\n\n\n\n

By using the property of geometric mean,<\/p>\n\n\n\n

b2<\/sup> = ac<\/p>\n\n\n\n

bc = ad<\/p>\n\n\n\n

c2<\/sup> = bd<\/p>\n\n\n\n

Let us consider RHS: (a + b)2<\/sup> + 2(b + c)2<\/sup> + (c + d)2<\/sup><\/p>\n\n\n\n

Let us expand<\/p>\n\n\n\n

(a + b)2<\/sup> + 2(b + c)2<\/sup> + (c + d)2<\/sup> = (a + b)2<\/sup> + 2 (a+b) (c+d) + (c+d)2<\/sup><\/p>\n\n\n\n

= a2<\/sup> + b2<\/sup> + 2ab + 2(c2<\/sup> + b2<\/sup> + 2cb) + c2<\/sup> + d2<\/sup> + 2cd<\/p>\n\n\n\n

= a2<\/sup> + b2<\/sup> + c2<\/sup> + d2<\/sup> + 2ab + 2(c2<\/sup> + b2<\/sup> + 2cb) + 2cd<\/p>\n\n\n\n

= a2<\/sup> + b2<\/sup> + c2<\/sup> + d2<\/sup> + 2(ab + bd + ac + cb +cd) [Since, c2<\/sup> = bd, b2<\/sup> = ac]<\/p>\n\n\n\n

You can visualize the above expression by making separate terms for (a + b + c)2<\/sup> + d2<\/sup> + 2d(a + b + c) = {(a + b + c) + d}2<\/sup><\/p>\n\n\n\n

\u2234 RHS = LHS<\/p>\n\n\n\n

Hence proved.<\/p>\n\n\n\n

(iii) <\/strong>(b + c) (b + d) = (c + a) (c + d)<\/p>\n\n\n\n

Given that a, b, c are in GP.<\/p>\n\n\n\n

By using the property of geometric mean,<\/p>\n\n\n\n

b2<\/sup> = ac<\/p>\n\n\n\n

bc = ad<\/p>\n\n\n\n

c2<\/sup> = bd<\/p>\n\n\n\n

Let us consider LHS: (b + c) (b + d)<\/p>\n\n\n\n

Upon expansion we get,<\/p>\n\n\n\n

(b + c) (b + d) = b2<\/sup> + bd + cb + cd<\/p>\n\n\n\n

= ac + c2<\/sup> + ad + cd [by using property of geometric mean]<\/p>\n\n\n\n

= c (a + c) + d (a + c)<\/p>\n\n\n\n

= (a + c) (c + d)<\/p>\n\n\n\n

= RHS<\/p>\n\n\n\n

\u2234 LHS = RHS<\/p>\n\n\n\n

Hence proved.<\/p>\n\n\n\n

10. If a, b, c are in G.P., prove that the following are also in G.P.:
(i) a2<\/sup>, b2<\/sup>, c2<\/sup><\/strong><\/p>\n\n\n\n

(ii) a3<\/sup>, b3<\/sup>, c3<\/sup><\/strong><\/p>\n\n\n\n

(iii) a2<\/sup> + b2<\/sup>, ab + bc, b2<\/sup> + c2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) <\/strong>a2<\/sup>, b2<\/sup>, c2<\/sup><\/p>\n\n\n\n

Given that a, b, c are in GP.<\/p>\n\n\n\n

By using the property of geometric mean,<\/p>\n\n\n\n

b2<\/sup> = ac<\/p>\n\n\n\n

on squaring both the sides we get,<\/p>\n\n\n\n

(b2<\/sup>)2<\/sup> = (ac)2<\/sup><\/p>\n\n\n\n

(b2<\/sup>)2<\/sup> = a2<\/sup>c2<\/sup><\/p>\n\n\n\n

\u2234 a2<\/sup>, b2<\/sup>, c2<\/sup> are in G.P.<\/p>\n\n\n\n

(ii) <\/strong>a3<\/sup>, b3<\/sup>, c3<\/sup><\/p>\n\n\n\n

Given that a, b, c are in GP.<\/p>\n\n\n\n

By using the property of geometric mean,<\/p>\n\n\n\n

b2<\/sup> = ac<\/p>\n\n\n\n

on squaring both the sides we get,<\/p>\n\n\n\n

(b2<\/sup>)3<\/sup> = (ac)3<\/sup><\/p>\n\n\n\n

(b2<\/sup>)3<\/sup> = a3<\/sup>c3<\/sup><\/p>\n\n\n\n

(b3<\/sup>)2<\/sup> = a3<\/sup>c3<\/sup><\/p>\n\n\n\n

\u2234 a3<\/sup>, b3<\/sup>, c3<\/sup> are in G.P.<\/p>\n\n\n\n

(iii) <\/strong>a2<\/sup> + b2<\/sup>, ab + bc, b2<\/sup> + c2<\/sup><\/p>\n\n\n\n

Given that a, b, c are in GP.<\/p>\n\n\n\n

By using the property of geometric mean,<\/p>\n\n\n\n

b2<\/sup> = ac<\/p>\n\n\n\n

a2<\/sup> + b2<\/sup>, ab + bc, b2<\/sup> + c2<\/sup> or (ab + bc)2<\/sup> = (a2<\/sup> + b2<\/sup>) (b2<\/sup> + c2<\/sup>) [by using the property of GM]<\/p>\n\n\n\n

Let us consider LHS: (ab + bc)2<\/sup><\/p>\n\n\n\n

Upon expansion we get,<\/p>\n\n\n\n

(ab + bc)2<\/sup> = a2<\/sup>b2<\/sup> + 2ab2<\/sup>c + b2<\/sup>c2<\/sup><\/p>\n\n\n\n

= a2<\/sup>b2<\/sup> + 2b2<\/sup>(b2<\/sup>) + b2<\/sup>c2<\/sup> [Since, ac = b2<\/sup>]<\/p>\n\n\n\n

= a2<\/sup>b2<\/sup> + 2b4<\/sup> + b2<\/sup>c2<\/sup><\/p>\n\n\n\n

= a2<\/sup>b2<\/sup> + b4<\/sup> + a2<\/sup>c2<\/sup> + b2<\/sup>c2<\/sup> {again using b2<\/sup> = ac }<\/p>\n\n\n\n

= b2<\/sup>(b2<\/sup> + a2<\/sup>) + c2<\/sup>(a2<\/sup> + b2<\/sup>)<\/p>\n\n\n\n

= (a2<\/sup> + b2<\/sup>)(b2<\/sup> + c2<\/sup>)<\/p>\n\n\n\n

= RHS<\/p>\n\n\n\n

\u2234 LHS = RHS<\/p>\n\n\n\n

Hence a2<\/sup> + b2<\/sup>, ab + bc, b2<\/sup> + c2<\/sup> are in GP.<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 20.6 PAGE NO: 20.54<\/p>\n\n\n\n

1. Insert 6 geometric means between 27 and 1\/81.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let the six terms be a1<\/sub>, a2<\/sub>, a3<\/sub>, a4<\/sub>, a5<\/sub>, a6<\/sub>.<\/p>\n\n\n\n

A = 27, B = 1\/81<\/p>\n\n\n\n

Now, these 6 terms are between A and B.<\/p>\n\n\n\n

So the GP is: A, a1<\/sub>, a2<\/sub>, a3<\/sub>, a4<\/sub>, a5<\/sub>, a6<\/sub>, B.<\/p>\n\n\n\n

So we now have 8 terms in GP with the first term being 27 and eighth being 1\/81.<\/p>\n\n\n\n

We know that, Tn<\/sub> = arn\u20131<\/sup><\/p>\n\n\n\n

Here, Tn<\/sub> = 1\/81, a = 27 and<\/p>\n\n\n\n

1\/81 = 27r8-1<\/sup><\/p>\n\n\n\n

1\/(81\u00d727) = r7<\/sup><\/p>\n\n\n\n

r = 1\/3<\/p>\n\n\n\n

a1<\/sub> = Ar = 27\u00d71\/3 = 9<\/p>\n\n\n\n

a2<\/sub> = Ar2<\/sup> = 27\u00d71\/9 = 3<\/p>\n\n\n\n

a3<\/sub> = Ar3<\/sup> = 27\u00d71\/27 = 1<\/p>\n\n\n\n

a4<\/sub> = Ar4<\/sup> = 27\u00d71\/81 = 1\/3<\/p>\n\n\n\n

a5<\/sub> = Ar5<\/sup> = 27\u00d71\/243 = 1\/9<\/p>\n\n\n\n

a6<\/sub> = Ar6<\/sup> = 27\u00d71\/729 = 1\/27<\/p>\n\n\n\n

\u2234 The six GM between 27 and 1\/81 are 9, 3, 1, 1\/3, 1\/9, 1\/27<\/p>\n\n\n\n

2. Insert 5 geometric means between 16 and 1\/4.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let the five terms be a1<\/sub>, a2<\/sub>, a3<\/sub>, a4<\/sub>, a5<\/sub>.<\/p>\n\n\n\n

A = 27, B = 1\/81<\/p>\n\n\n\n

Now, these 5 terms are between A and B.<\/p>\n\n\n\n

So the GP is: A, a1<\/sub>, a2<\/sub>, a3<\/sub>, a4<\/sub>, a5<\/sub>, B.<\/p>\n\n\n\n

So we now have 7 terms in GP with the first term being 16 and seventh being 1\/4.<\/p>\n\n\n\n

We know that, Tn<\/sub> = arn\u20131<\/sup><\/p>\n\n\n\n

Here, Tn<\/sub> = 1\/4, a = 16 and<\/p>\n\n\n\n

1\/4 = 16r7-1<\/sup><\/p>\n\n\n\n

1\/(4\u00d716) = r6<\/sup><\/p>\n\n\n\n

r = 1\/2<\/p>\n\n\n\n

a1<\/sub> = Ar = 16\u00d71\/2 = 8<\/p>\n\n\n\n

a2<\/sub> = Ar2<\/sup> = 16\u00d71\/4 = 4<\/p>\n\n\n\n

a3<\/sub> = Ar3<\/sup> = 16\u00d71\/8 = 2<\/p>\n\n\n\n

a4<\/sub> = Ar4<\/sup> = 16\u00d71\/16 = 1<\/p>\n\n\n\n

a5<\/sub> = Ar5<\/sup> = 16\u00d71\/32 = 1\/2<\/p>\n\n\n\n

\u2234 The five GM between 16 and 1\/4 are 8, 4, 2, 1, \u00bd<\/p>\n\n\n\n

3. Insert 5 geometric means between 32\/9 and 81\/2.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let the five terms be a1<\/sub>, a2<\/sub>, a3<\/sub>, a4<\/sub>, a5<\/sub>.<\/p>\n\n\n\n

A = 32\/9, B = 81\/2<\/p>\n\n\n\n

Now, these 5 terms are between A and B.<\/p>\n\n\n\n

So the GP is: A, a1<\/sub>, a2<\/sub>, a3<\/sub>, a4<\/sub>, a5<\/sub>, B.<\/p>\n\n\n\n

So we now have 7 terms in GP with the first term being 32\/9 and seventh being 81\/2.<\/p>\n\n\n\n

We know that, Tn<\/sub> = arn\u20131<\/sup><\/p>\n\n\n\n

Here, Tn<\/sub> = 81\/2, a = 32\/9 and<\/p>\n\n\n\n

81\/2 = 32\/9r7-1<\/sup><\/p>\n\n\n\n

(81\u00d79)\/(2\u00d732) = r6<\/sup><\/p>\n\n\n\n

r = 3\/2<\/p>\n\n\n\n

a1<\/sub> = Ar = (32\/9)\u00d73\/2 = 16\/3<\/p>\n\n\n\n

a2<\/sub> = Ar2<\/sup> = (32\/9)\u00d79\/4 = 8<\/p>\n\n\n\n

a3<\/sub> = Ar3<\/sup> = (32\/9)\u00d727\/8 = 12<\/p>\n\n\n\n

a4<\/sub> = Ar4<\/sup> = (32\/9)\u00d781\/16 = 18<\/p>\n\n\n\n

a5<\/sub> = Ar5<\/sup> = (32\/9)\u00d7243\/32 = 27<\/p>\n\n\n\n

\u2234 The five GM between 32\/9 and 81\/2 are 16\/3, 8, 12, 18, 27<\/p>\n\n\n\n

4. Find the geometric means of the following pairs of numbers:
(i) 2 and 8
(ii) a3<\/sup>b and ab3<\/sup>
(iii) \u20138 and \u20132<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i)<\/strong> 2 and 8<\/p>\n\n\n\n

GM between a and b is \u221aab<\/p>\n\n\n\n

Let a = 2 and b =8<\/p>\n\n\n\n

GM = \u221a2\u00d78<\/p>\n\n\n\n

= \u221a16<\/p>\n\n\n\n

= 4<\/p>\n\n\n\n

(ii)<\/strong> a3<\/sup>b and ab3<\/sup><\/p>\n\n\n\n

GM between a and b is \u221aab<\/p>\n\n\n\n

Let a = a3<\/sup>b and b = ab3<\/sup><\/p>\n\n\n\n

GM = \u221a(a3<\/sup>b \u00d7 ab3<\/sup>)<\/p>\n\n\n\n

= \u221aa4<\/sup>b4<\/sup><\/p>\n\n\n\n

= a2<\/sup>b2<\/sup><\/p>\n\n\n\n

(iii)<\/strong> \u20138 and \u20132<\/p>\n\n\n\n

GM between a and b is \u221aab<\/p>\n\n\n\n

Let a = \u20132 and b = \u20138<\/p>\n\n\n\n

GM = \u221a(\u20132\u00d7\u20138)<\/p>\n\n\n\n

= \u221a\u201316<\/p>\n\n\n\n

= -4<\/p>\n\n\n\n

5. If a is the G.M. of 2 and \u00bc find a.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know that GM between a and b is \u221aab<\/p>\n\n\n\n

Let a = 2 and b = 1\/4<\/p>\n\n\n\n

GM = \u221a(2\u00d71\/4)<\/p>\n\n\n\n

= \u221a(1\/2)<\/p>\n\n\n\n

= 1\/\u221a2<\/p>\n\n\n\n

\u2234 value of a is 1\/\u221a2<\/p>\n\n\n\n

6. Find the two numbers whose A.M. is 25 and GM is 20.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given: A.M = 25, G.M = 20.<\/p>\n\n\n\n

G.M = \u221aab<\/p>\n\n\n\n

A.M = (a+b)\/2<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

\u221aab = 20 \u2026\u2026. (1)<\/p>\n\n\n\n

(a+b)\/2 = 25\u2026\u2026. (2)<\/p>\n\n\n\n

a + b = 50<\/p>\n\n\n\n

a = 50 \u2013 b<\/p>\n\n\n\n

Putting the value of \u2018a\u2019 in equation (1), we get,<\/p>\n\n\n\n

\u221a[(50-b)b] = 20<\/p>\n\n\n\n

50b \u2013 b2<\/sup> = 400<\/p>\n\n\n\n

b2<\/sup> \u2013 50b + 400 = 0<\/p>\n\n\n\n

b2<\/sup> \u2013 40b \u2013 10b + 400 = 0<\/p>\n\n\n\n

b(b \u2013 40) \u2013 10(b \u2013 40) = 0<\/p>\n\n\n\n

b = 40 or b = 10<\/p>\n\n\n\n

If b = 40 then a = 10<\/p>\n\n\n\n

If b = 10 then a = 40<\/p>\n\n\n\n

\u2234 The numbers are 10 and 40.<\/p>\n\n\n\n

7.<\/strong> Construct a quadratic in x such that A.M. of its roots is A and G.M. is G.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let the root of the quadratic equation be a and b.<\/p>\n\n\n\n

So, according to the given condition,<\/p>\n\n\n\n

A.M = (a+b)\/2 = A<\/p>\n\n\n\n

a + b = 2A \u2026.. (1)<\/p>\n\n\n\n

GM = \u221aab = G<\/p>\n\n\n\n

ab = G2<\/sup>\u2026 (2)<\/p>\n\n\n\n

The quadratic equation is given by,<\/p>\n\n\n\n

x2 <\/sup>\u2013 x (Sum of roots) + (Product of roots) = 0<\/p>\n\n\n\n

x2<\/sup> \u2013 x (2A) + (G2<\/sup>) = 0<\/p>\n\n\n\n

x2<\/sup> \u2013 2Ax + G2<\/sup> = 0 [Using (1) and (2)]<\/p>\n\n\n\n

\u2234 The required quadratic equation is x2<\/sup> \u2013 2Ax + G2<\/sup> = 0.<\/p>\n\n\n\n

RD Sharma Solutions for Class 11 Maths Chapter 20: Download PDF<\/strong><\/h2>\n\n\n\n

RD Sharma Solutions for Class 11 Maths Chapter 20\u2013Geometric Progressions<\/p>\n\n\n\n

Download PDF<\/strong>: RD Sharma Solutions for Class 11 Maths Chapter 20\u2013Geometric Progressions PDF<\/a><\/p>\n\n\n\n

Chapterwise RD Sharma Solutions for Class 11 Maths :<\/strong><\/h2>\n\n\n\n