{"id":565585,"date":"2021-12-28T10:42:41","date_gmt":"2021-12-28T10:42:41","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=565585"},"modified":"2021-12-29T11:07:51","modified_gmt":"2021-12-29T11:07:51","slug":"maharashtra-board-solutions-for-class-10-maths-part-1-chapter-3-arithmetic-progression","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-for-class-10-maths-part-1-chapter-3-arithmetic-progression\/","title":{"rendered":"Maharashtra Board Solutions for Class 10-Maths (Part 1): Chapter 3- Arithmetic Progression"},"content":{"rendered":"\n

Class 10: Maths Chapter 3 solutions. Complete Class 10 Maths Chapter 3 Notes.<\/p>\n\n\n\n

Maharashtra Board Solutions for Class 10-Maths (Part 1): Chapter 3- Arithmetic Progression<\/h2>\n\n\n\n

Maharashtra Board 10th Maths Chapter 3, Class 10 Maths Chapter 3 solutions<\/p>\n\n\n\n

Practice Set 3.1<\/h4>\n\n\n\n

Question 1.
Which of the following sequences are A.P.? If they are A.P. find the common difference.<\/strong>
\"Maharashtra
Solution:
<\/strong>i. The given sequence is 2, 4, 6, 8,\u2026
Here, t1<\/sub> = 2, t2<\/sub> = 4, t3<\/sub> = 6, t4<\/sub> = 8
\u2234 t2<\/sub> \u2013 t1<\/sub> = 4 \u2013 2 = 2
t3<\/sub> \u2013 t2<\/sub> = 6 \u2013 4 = 2
t4<\/sub> \u2013 t3<\/sub> = 8 \u2013 6 = 2
\u2234 t2<\/sub> \u2013 t1<\/sub> =  t3<\/sub> \u2013 t2<\/sub> = \u2026 = 2 = d = constant
The difference between two consecutive terms is constant.
\u2234 The given sequence is an A.P. and common difference (d) = 2.<\/p>\n\n\n\n

\"Maharashtra
\"Maharashtra
The difference between two consecutive terms is constant.
\u2234 The given sequence is an A.P. and common difference (d) = 12.<\/p>\n\n\n\n

iii. The given sequence is -10, -6, -2, 2,\u2026
Here, t1<\/sub> = -10, t2<\/sub> = \u2013 6, t3<\/sub> = -2, t4<\/sub> = 2
\u2234 t2<\/sub> \u2013 t1<\/sub> = -6 \u2013 (-10) = -6 + 10 = 4
t3<\/sub> \u2013 t2<\/sub> = -2 -(-6) = -2 + 6 = 4
t4<\/sub> \u2013 t3<\/sub> = 2 \u2013 (-2) = 2 + 2 = 4
\u2234 t2<\/sub> \u2013 t1<\/sub> = t3<\/sub> \u2013 t2<\/sub> = \u2026 = 4 = d = constant
The difference between two consecutive terms is constant.
\u2234 The given sequence is an A.P. and common difference (d) = 4.<\/p>\n\n\n\n

iv. The given sequence is 0.3, 0.33, 0.333,\u2026
Here, t1<\/sub> = 0.3, t2<\/sub> = 0.33, t3<\/sub> = 0.333
\u2234 t2<\/sub> -t1<\/sub> = 0.33 \u2013 0.3 = 0.03
t3<\/sub> \u2013 t2<\/sub> = 0.333 \u2013 0.33 = 0.003
\u2234 t2<\/sub> \u2013 t1<\/sub> \u2260 t3<\/sub> \u2013 t2<\/sub>
The difference between two consecutive terms is not constant.
\u2234 The given sequence is not an A.P.<\/p>\n\n\n\n

v. The given sequence is 0, -4, -8, -12,\u2026
Here, t1<\/sub> = 0, t2<\/sub> = -4, t3<\/sub> = -8, t4<\/sub> = -12
\u2234 t2<\/sub> \u2013 t1<\/sub> = -4 \u2013 0 = -4
t3<\/sub> \u2013 t2<\/sub> = -8 \u2013 (-4) = -8 + 4 = -4
t4<\/sub> \u2013 t3<\/sub> = -12 \u2013 (-8) = -12 + 8 = -4
\u2234 t2<\/sub> \u2013 t1<\/sub> = t3<\/sub> \u2013 t2<\/sub> = \u2026 = \u20144 = d = constant
The difference between two consecutive terms is constant.
\u2234 The given sequence is an A.P. and common difference (d) = -4.<\/p>\n\n\n\n

\"Maharashtra
The difference between two consecutive terms is constant.
\u2234 The given sequence is an A.P. and common difference (d) = 0.<\/p>\n\n\n\n

\"Maharashtra
The difference between two consecutive terms is constant.
\u2234 The given sequence is an A.P. and common difference (d) = \u221a2.<\/p>\n\n\n\n

viii. The given sequence is 127, 132, 137,\u2026
Here, t1<\/sub> = 127, t2<\/sub> = 132, t3<\/sub> = 137
\u2234 t2<\/sub> \u2013 t1<\/sub> = 132 \u2013 127 = 5
t3<\/sub> \u2013 t2 <\/sub>= 137 \u2013 132 = 5
\u2234 t2<\/sub> \u2013 t1<\/sub> = t3<\/sub> \u2013 t2<\/sub> = \u2026 = 5 = d = constant
The difference between two consecutive terms is constant.
\u2234 The given sequence is an A.P. and common difference (d) = 5.<\/p>\n\n\n\n

Question 2.
Write an A.P. whose first term is a and common difference is d in each of the following.<\/strong>
i. a = 10, d = 5
ii. a = -3, d = 0
iii. a = -7, d = 12
iv. a = -1.25, d = 3
v. a = 6, d = -3
vi. a = -19, d = -4
Solution:
<\/strong>i. a = 10, d = 5 \u2026[Given]
\u2234 t1<\/sub> = a = 10
t2<\/sub> = t1<\/sub> + d = 10 + 5 = 15
t3<\/sub> = t2<\/sub> + d = 15 + 5 = 20
t4<\/sub> = t3<\/sub> + d = 20 + 5 = 25
\u2234 The required A.P. is 10,15, 20, 25,\u2026<\/p>\n\n\n\n

ii. a = -3, d = 0 \u2026[Given]
\u2234 t1<\/sub> = a = -3
t2<\/sub> = t1<\/sub> + d = -3 + 0 = -3
t3<\/sub> = t2<\/sub> + d = -3 + 0 = -3
t4<\/sub> = t3<\/sub> + d = -3 + 0 = -3
\u2234 The required A.P. is -3, -3, -3, -3,\u2026<\/p>\n\n\n\n

\"Maharashtra
\"Maharashtra
\u2234 The required A.P. is -7, \u2013 6.5, \u2013 6, \u2013 5.5,<\/p>\n\n\n\n

iv. a = -1.25, d = 3 \u2026[Given]
t1<\/sub> = a = -1.25
t2<\/sub> = t1<\/sub> + d = \u2013 1.25 + 3 = 1.75
t3<\/sub> = t2<\/sub> + d = 1.75 + 3 = .4.75
t4<\/sub> = t3<\/sub> + d = 4.75 + 3 = 7.75
\u2234 The required A.P. is -1.25, 1.75, 4.75, 7.75,\u2026<\/p>\n\n\n\n

v. a = 6, d = -3 \u2026[Given]
\u2234 t1<\/sub> = a = 6
t2<\/sub> = t1<\/sub> + d = 6 \u2013 3 = 3
t3<\/sub> = t2<\/sub> + d = 3 \u2013 3 = 0
t4<\/sub> = t3<\/sub> + d = 0- 3 = -3
\u2234 The required A.P. is 6, 3, 0, -3,\u2026<\/p>\n\n\n\n

vi. a = -19, d = -4 \u2026[Given]
t1<\/sub> = a = -19
t2<\/sub> = t1<\/sub> + d = -19 \u2013 4 = -23
t3<\/sub> = t2<\/sub> + d = -23 \u2013 4 = -27
t4<\/sub> = t3<\/sub> + d = -27 \u2013 4 = -31
\u2234 The required A.P. is -19, -23, -27, -31,\u2026<\/p>\n\n\n\n

Question 3.
Find the first term and common difference for each of the A.P.<\/strong>
\"Maharashtra
Solution:
<\/strong>i. The given A.P. is 5, 1,-3,-7,\u2026
Here, t1<\/sub> = 5, t2<\/sub> = 1
\u2234 a = t1<\/sub> = 5 and
d = t2<\/sub> \u2013 t1<\/sub> = 1 \u2013 5 = -4
\u2234 first term (a) = 5,
common difference (d) = -4<\/p>\n\n\n\n

ii. The given A.P. is 0.6, 0.9, 1.2, 1.5,\u2026
Here, t1<\/sub> = 0.6, t2<\/sub> = 0.9
\u2234 a = t1<\/sub> = 0.6 and
d = t2<\/sub> \u2013 t1 <\/sub>= 0.9 \u2013 0.6 = 0.3
\u2234 first term (a) = 0.6,
common difference (d) = 0.3<\/p>\n\n\n\n

iii. The given A.P. is 127, 135, 143, 151,\u2026
Here, t1<\/sub> = 127, t2<\/sub> = 135
\u2234 a = t1<\/sub> = 127 and
d = t2<\/sub> \u2013 t1 <\/sub>= 135 \u2013 127 = 8
\u2234 first term (a) = 127,
common difference (d) = 8<\/p>\n\n\n\n

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

Question 1.
Complete the given pattern. Look at the pattern of the numbers. Try to find a rule to obtain the next number from its preceding number. Write the next numbers. (Textbook pg, no. 55 and 56)
Answer:<\/strong>
\"Maharashtra
Every pattern is formed by adding a circle in horizontal and vertical rows to the preceding pattern.
\u2234 The sequence for the above pattern is 1,3, 5, 7, 9,11,13,15,17,\u2026.
\"Maharashtra
Every pattern is formed by adding 2 triangles horizontally and 1 triangle vertically to the preceding pattern.
\u2234 The sequence for the above pattern is 5,8,11,14,17,20,23,\u2026<\/p>\n\n\n\n

Question 2.
Some sequences are given below. Show the positions of the terms by t1<\/sub>, t2<\/sub>, t3<\/sub>,\u2026
Answer:<\/strong>
\"Maharashtra<\/p>\n\n\n\n

Question 3.
Some sequences are given below. Check whether there is any rule among the terms. Find the similarity between two sequences. To check the rule for the terms of the sequence look at the arrangements and fill the empty boxes suitably. (Textbook pg. no. 56 and 57)<\/strong>
i. .1,4,7,10,13,\u2026
ii. 6,12,18,24,\u2026
iii. 3,3,3,3,\u2026
iv. 4, 16, 64,\u2026
v. -1, -1.5, -2, -2.5,\u2026
vi. 13<\/sup>, 23<\/sup>, 33<\/sup>, 43<\/sup>
Answer:
<\/strong>\"Maharashtra
\"Maharashtra
The similarity in the sequences i., ii., iii. and v. is that the next term is obtained by adding a particular fixed number to the previous term.<\/p>\n\n\n\n

Note : A Geometric Progression is a sequence in which the ratio of any two consecutive terms is a constant,
\"Maharashtra
Sequence iv. is a geometric progression.<\/p>\n\n\n\n

Question 4.
Write one example of finite and infinite A.P. each. (Textbook pg. no. 59)
Answer:<\/strong>
Finite A.P.:
Even natural numbers from 4 to 50:
4, 6, 8, \u2026\u2026\u2026\u2026\u2026\u2026. 50.
Infinite A. P.:
Positive multiples of 5:
5, 10, 15, \u2026\u2026\u2026\u2026\u2026..<\/p>\n\n\n\n

Practice Set 3.2<\/h4>\n\n\n\n

Question 1.
Write the correct number in the given boxes from the following A.P.<\/strong>
\"Maharashtra
Solution:
<\/strong>\"Maharashtra<\/p>\n\n\n\n

Question 2.
Decide whether following sequence is an A.P., if so find the 20th<\/sup> term of the progression.
-12, -5, 2, 9,16, 23,30,\u2026
Solution:<\/strong>
i. The given sequence is
-12, -5,2, 9, 16, 23,30,\u2026
Here, t1<\/sub> = -12<\/sub>, t2<\/sub> = -5, t3<\/sub> = 2, t4<\/sub> = 9
\u2234 t2<\/sub> \u2013 t1<\/sub> \u2013 5 \u2013 (-12) \u2013 5 + 12 = 7
t3<\/sub> \u2013 t2<\/sub> = 2 \u2013 (-5) = 2 + 5 = 7
\u2234 t4<\/sub> \u2013 t3<\/sub> \u2013 9 \u2013 2 = 7
\u2234 t2<\/sub> \u2013 t1<\/sub> = t3<\/sub> \u2013 t2<\/sub> = \u2026 = 7 = d = constant
The difference between two consecutive terms is constant.
\u2234 The given sequence is an A.P.<\/p>\n\n\n\n

ii. tn<\/sub> = a + (n \u2013 1)d
\u2234 t20<\/sub> = -12 + (20 \u2013 1)7 \u2026[\u2235a = -12, d = 7]
= -12 + 19 \u00d7 7
= -12 + 133
\u2234 t20<\/sub> = 121
\u2234 20th<\/sup> term of the given A.P. is 121.<\/p>\n\n\n\n

Question 3.
Given Arithmetic Progression is 12, 16, 20, 24, \u2026 Find the 24th<\/sup> term of this progression.
Solution:<\/strong>
The given A.P. is 12, 16, 20, 24,\u2026
Here, a = 12, d = 16 \u2013 12 = 4 Since,
tn<\/sub> = a + (n \u2013 1)d
\u2234 t24<\/sub> = 12 + (24 \u2013 1)4
= 12 + 23 \u00d7 4
= 12 + 92
\u2234 t24<\/sub> = 104
\u2234 24th term of the given A.P. is 104.<\/p>\n\n\n\n

Question 4.
Find the 19th term of the following A.P. 7,13,19,25\u2026..
Solution:<\/strong>
The given A.P. is 7, 13, 19, 25,\u2026
Here, a = 7, d = 13 \u2013 7 = 6
Since, tn<\/sub> = a + (n \u2013 1)d
\u2234 t19<\/sub> = 7 + (19 \u2013 1)6
= 7 + 18 \u00d7 6
= 7 + 108
\u2234 t19<\/sub> = 115
\u2234 19th<\/sup> term of the given A.P. is 115.<\/p>\n\n\n\n

Question 5.
Find the 27th<\/sup> term of the following A.P. 9,4,-1,-6,-11,\u2026
Solution:<\/strong>
The given A.P. is 9, 4, -1, -6, -11,\u2026
Here, a = 9, d = 4- 9 = -5
Since, tn<\/sub> = a + (n \u2013 1)d
\u2234 t27<\/sub> = 9 + (27 \u2013 1)(-5)
= 9 + 26 \u00d7 (-5)
= 9 \u2013 130
\u2234 t27<\/sub> = -121
\u2234 27th<\/sup> term of the given A.P. is -121.<\/p>\n\n\n\n

Question 6.
Find how many three digit natural numbers are divisible by 5.
Solution:<\/strong>
The three digit natural numbers divisible by
5 are 100, 105, 110, \u2026,995
The above sequence is an A.P.
\u2234 a = 100, d = 105 \u2013 100 = 5
Let the number of terms in the A.P. be n.
Then, tn<\/sub> = 995
Since, tn<\/sub> = a + (n \u2013 1)d
\u2234 995 = 100 +(n \u2013 1)5
\u2234 995 \u2013 100 = (n \u2013 1)5
\u2234 895 = (n \u2013 1)5
\u2234 n \u2013 1 = 8955
\u2234 n \u2013 1 = 179
\u2234 n = 179 + 1 = 180
\u2234 There are 180 three digit natural numbers which are divisible by 5.<\/p>\n\n\n\n

Question 7.
The 11th term and the 21st term of an A.P. are 16 and 29 respectively, then find the 41st term of that A.P.
Solution:<\/strong>
Bor an A.P., let a be the first term and d be the common difference,
t11<\/sub> = 16, t21<\/sub> = 29 \u2026[Given]
tn<\/sub> = a + (n \u2013 1)d
\u2234 t11<\/sub>, = a + (11 \u2013 1)d
\u2234 16 = a + 10d
i.e. a + 10d = 16 \u2026(i)
Also, t21<\/sub> = a + (21 \u2013 1)d
\u2234 29 = a + 20d
i.e. a + 20d = 29 \u2026(ii)
Subtracting equation (i) from (ii), we get a
\"Maharashtra<\/p>\n\n\n\n

Question 8.
8. 11, 8, 5, 2, \u2026 In this A.P. which term is number-151?
Solution:<\/strong>
The given A.P. is 11, 8, 5, 2,\u2026
Here, a = 11, d = 8 \u2013 11 = -3
Let the nth<\/sup> term of the given A.P. be -151.
Then, tn<\/sub> = \u2013 151
Since, tn<\/sub> = a + (n \u2013 1)d
\u2234 -151= 11 + (n \u2013 1)(-3)
\u2234 -151 \u2013 11 =(n \u2013 1)(-3)
\u2234 -162 = (n \u2013 1)(-3)
\u2234 n \u2013 1 = \u2212162\u22123
\u2234 n \u2013 1 = 54
\u2234 n = 54 + 1 = 55
\u2234 55th term of the given A.P. is -151.<\/p>\n\n\n\n

Question 9.
In the natural numbers from 10 to 250, how many are divisible by 4?
Solution:<\/strong>
The natural numbers from 10 to 250 divisible
by 4 are 12, 16, 20, \u2026,248
The above sequence is an A.P.
\u2234 a = 12, d = 16 \u2013 12 = 4
Let the number of terms in the A.P. be n.
Then, tn<\/sub> = 248
Since, tn<\/sub> = a + (n \u2013 1)d
\u2234 248 = 12 + (n \u2013 1)4
\u2234 248 \u2013 12 = (n \u2013 1)4
\u2234 236 = (n \u2013 1)4
\u2234 n \u2013 1 = 2364
\u2234 n \u2013 1 = 59
\u2234 n = 59 + 1 = 60
\u2234 There are 60 natural numbers from 10 to 250 which are divisible by 4.<\/p>\n\n\n\n

Question 10.
In an A.P. 17th<\/sup> term is 7 more than its 10th term. Find the common difference.
Solution:<\/strong>
For an A.P., let a be the first term and d be the common difference.
According to the given condition,
t17<\/sub> = t10<\/sub> + 7
\u2234 a + (17 \u2013 1)d = a + (10 \u2013 1)d + 7 \u2026[\u2235 tn<\/sub> = a + (n \u2013 1)d]
\u2234 a + 16d = a + 9d + 7
\u2234 a + 16d \u2013 a \u2013 9d = 7
\u2234 7d = 7
\u2234 d = 77 = 1
\u2234 The common difference is 1.<\/p>\n\n\n\n

Question 1.
Kabir\u2019s mother keeps a record of his height on each birthday. When he was one year old, his height was 70 cm, at 2 years he was 80 cm tall and 3 years he was 90 cm tall. His aunt Meera was studying in the 10th class. She said, \u201cit seems like Kabir\u2019s height grows in Arithmetic Progression\u201d. Assuming this, she calculated how tall Kabir will be at the age of 15 years when he is in 10th! She was shocked to find it. You too assume that Kabir grows in A.P. and find out his height at the age of 15 years. (Textbook pg. no. 63)
Solution:<\/strong>
Height of Kabir when he was 1 year old = 70 cm Height of Kabir when he was 2 years old = 80 cm
Height of Kabir when he was 3 years old = 90 cm The heights of Kabir form an A.P.
Here, a = 70, d = 80 \u2013 70 = 10
We have to find height of Kabir at the age of 15years i.e. t15<\/sub>.
Now, tn = a + (n \u2013 1)d
\u2234 t15<\/sub> = 70 + (15 \u2013 1)10
= 70 + 14 \u00d7 10 = 70 + 140
\u2234 t15<\/sub> = 210
\u2234 The height of Kabir at the age of 15 years will be 210 cm.<\/p>\n\n\n\n

Question 2.
Is 5, 8, 11, 14, \u2026. an A.P.? If so then what will be the 100th<\/sup> term? Check whether 92 is in this A.P.? Is number 61 in this A.P.? (Textbook pg. no, 62)
Solution:<\/strong>
i. The given sequence is
5, 8,11,14,\u2026
Here, t1<\/sub> = 5, t2<\/sub> = 8, t3<\/sub> = 11, t4<\/sub> = 14
\u2234 t2<\/sub> \u2013 t1<\/sub> = 8 \u2013 5 = 3
t3<\/sub> \u2013 t2 <\/sub>= 11 \u2013 8 = 3
t4<\/sub> \u2013 t3<\/sub> = 14 \u2013 11 = 3
\u2234 t2<\/sub> \u2013 t1<\/sub> = t3<\/sub> \u2013 t2<\/sub> = t4<\/sub> \u2013 t3<\/sub> = 3 = d = constant
The difference between two consecutive terms is constant
\u2234 The given sequence is an A.P.<\/p>\n\n\n\n

ii. tn<\/sub> = a + (n \u2013 1)d
\u2234 t100<\/sub> = 5 + (100 \u2013 1)3 \u2026[\u2235 a = 5, d = 3]
= 5 + 99 \u00d7 3
= 5 + 297
\u2234 t100<\/sub> = 302
\u2234 100th<\/sup> term of the given A.P. is 302.<\/p>\n\n\n\n

iii. To check whether 92 is in given A.P., let tn<\/sub> = 92
\u2234 tn<\/sub> = a + (n \u2013 1)d
\u2234 92 = 5 + (n \u2013 1)3
\u2234 92 = 5 + 3n \u2013 3
\u2234 92 = 2 + 3n
\u2234 90 = 3n
\u2234 n = 903 = 30
\u2234 92 is the 30th term of given A.P.<\/p>\n\n\n\n

iv. To check whether 61 is in given A.P., let tn<\/sub> = 61
61 = 5 + (n \u2013 1)3
\u2234 61 = 5 + 3n \u2013 3
\u2234 61 = 2 + 3n
\u2234 61 \u2013 2 = 3n
\u2234 59 = 3n
\u2234 n = 593
But, n is natural number 59
\u2234 n \u2260 593
\u2234 61 is not in given A.P.<\/p>\n\n\n\n

Practice Set 3.3<\/h4>\n\n\n\n

First term and common difference of an A.P. are 6 and 3 respectively; find S27<\/sub>.
Solution:<\/strong>
\"Maharashtra<\/p>\n\n\n\n

Find the sum of first 123 even natural numbers.
Solution:<\/strong>
The even natural numbers are 2, 4, 6, 8,\u2026
The above sequence is an A.P.
\u2234 a = 2, d = 4 \u2013 2 = 2, n = 123
\"Maharashtra
\u2234 The sum of first 123 even natural numbers is 15252.<\/p>\n\n\n\n

Find the sum of all even numbers between 1 and 350.
Solution:<\/strong>
The even numbers between 1 and 350 are 2, 4, 6,\u2026, 348.
The above sequence is an A.P.
\u2234 a = 2, d = 4 \u2013 2 = 2, tn<\/sub> = 348
Since, tn<\/sub> = a + (n \u2013 1)d
\u2234 348 = 2 + (n \u2013 1)2
\u2234 348 \u2013 2 = (n \u2013 1)2
\u2234 346 = (n \u2013 1)2
\u2234 n \u2013 1 = 3462
\u2234 n \u2013 1 = 173
\u2234 n = 173 + 1 = 174
Now, Sn<\/sub> = n2 [2a + (n \u2013 1)d]
\u2234 S174<\/sub> = 1742 [2 (2) + (174 \u2013 1)2]
= 87(4 + 173 \u00d7 2)
= 87(4 + 346)
= 87 \u00d7 350
\u2234 S174<\/sub> = 30450
\u2234 The sum of all even numbers between 1 and 350 is 30450.<\/p>\n\n\n\n

In an A.P. 19th<\/sup> term is 52 and 38th<\/sup> term is 128, find sum of first 56 terms.
Solution:<\/strong>
For an A.P., let a be the first term and d be the common difference.
t19<\/sub> = 52, t38<\/sub> = 128 \u2026[Given]
Since, tn<\/sub> = a + (n \u2013 1)d
\u2234 t19<\/sub> = a + (19 \u2013 1)d
\u2234 52 = a + 18d
i.e. a + 18d = 52 \u2026(i)
Also, t38<\/sub> = a + (38 \u2013 1)d
\u2234 128 = a + 37d
i.e. a + 37d = 128 \u2026(ii)
Adding equations (i) and (ii), we get
\"Maharashtra<\/p>\n\n\n\n

Complete the following activity to find the sum of natural numbers between 1 to 140 which are divisible by 4.
Solution:<\/strong>
\"Maharashtra<\/p>\n\n\n\n

Sum of first 55 terms in an A.P. is 3300, find its 28th<\/sup> term.
Solution:<\/strong>
For an A.P., let a be the first term and d be the common difference.
S55<\/sub> =3300 \u2026[Given]
Since, Sn = n2 [2a + (n \u2013 1)d]
\u2234 S55<\/sub> = 552 [2a + (55 \u2013 1)d]
\"Maharashtra<\/p>\n\n\n\n

Arithmetic Practice Set Question 7.
<\/strong>In an A.P. sum of three consecutive terms is 27 and their product is 504, find the terms. (Assume that three consecutive terms in A.P. are a \u2013 d, a, a + d.)
Solution:<\/strong>
Let the three consecutive terms in an A.P. be
a \u2013 d, a and a + d.
According to the first condition,
a \u2013 d + a + a + d = 27
\u2234 3a = 27
\u2234 a = 273
\u2234 a = 9 \u2026.(i)
According to the second condition,
(a \u2013 d) a (a + d) = 504
\u2234 a(a2<\/sup> \u2013 d2<\/sup>) = 504
\u2234 9(a2<\/sup> \u2013 d2<\/sup>) = 504 \u2026[From (i)]
\u2234 9(81 \u2013 d2<\/sup>) = 504
\u2234 81 \u2013 d2<\/sup> = 5049
\u2234 81 \u2013 d2<\/sup> = 56
\u2234 d2<\/sup> = 81 \u2013 56
\u2234 d2<\/sup> = 25
Taking square root of both sides, we get
d = \u00b1 5
When d = 5 and a =9,
a \u2013 d 9 \u2013 5 = 4
a = 9
a + d 9 + 5 = 14
When d = -5 and a = 9,
a \u2013 d = 9 \u2013 (-5) = 9 + 5 = 14
a = 9
a + d = 9 \u2013 5 = 4
\u2234 The three consecutive terms are 4, 9 and 14 or 14, 9 and 4.<\/p>\n\n\n\n

10th Maths 1 Practice Set 3.3 Question 8.
Find four consecutive terms in an A.P. whose sum is 12 and sum of 3rd<\/sup> and 4th<\/sup> term is 14. (Assume the four consecutive terms in A.P. are a \u2013 d, a, a + d, a + 2d.)
Solution:<\/strong>
Let the four consecutive terms in an A.P. be
a \u2013 d, a, a + d and a + 2d.
According to the first condition,
a \u2013 d + a + a + d + a + 2d = 12
\u2234 4a + 2d =12
\u2234 2(2a + d) = 12
\u2234 2a + d = 122
\u2234 2a + d = 6 \u2026(i)
According to the second condition,
a + d + a + 2d = 14
\u2234 2a + 3d = 14 \u2026(ii)
Subtracting equation (i) from (ii), we get
\"Maharashtra
\u2234 The four consecutive terms are -3,1,5 and 9.<\/p>\n\n\n\n

Math 1 Practice Set 3.3 Question 9.
<\/strong>If the 9th<\/sup> term of an A.P. is zero, then show that the 29th<\/sup> term is twice the 19th term.
To prove: t29<\/sub> = 2t19<\/sub>
Proof:<\/strong>
For an A.P., let a be the first term and d be the common difference.
t9<\/sub> = 0 \u2026[Given]
Since, tn = a + (n \u2013 1)d
\u2234 t9<\/sub> = a + (9 \u2013 1)d
\u2234 0 = a + 8d
\u2234 a = -8d \u2026(i)
Also, t19 <\/sub>= a + (19 \u2013 1)d
= a + 18d
= -8d + 18d \u2026 [From (i)]
\u2234 t19<\/sub> = 10d \u2026(ii)
and t29<\/sub> = a + (29 \u2013 1)d
= a + 28d
= -8d + 28d \u2026[From (i)]
\u2234 t29<\/sub> = 20d = 2(10d)
\u2234 t29<\/sub> = 2(t19<\/sub>) \u2026 [From (ii)]
\u2234 The 29th<\/sup> term is twice the 19th term.<\/p>\n\n\n\n

10 Class Math Part 1 Practice Set 3.3 Question 1.
Find the sum of all odd numbers from 1 to 150. (Textbook pg, no. 71)
Solution:<\/strong>
Odd numbers from 1 to 150 are 1,3, 5, 7,\u2026, 149
Here, difference between any two consecutive terms is 2.
\u2234 It is an A.P.
\u2234 a = 1, d = 2
Let us find how many odd numbers are there from 1 to 150, i.e. find the value of n if
tn<\/sub> = 149
tn<\/sub> = a + (n \u2013 1)d
\u2234 149 = 1 + (n \u2013 1)2
\u2234 149 \u2013 1 = (n \u2013 1)2
\u2234 1482 = n \u2013 1
\u2234 74 = n \u2013 1
\u2234 n = 74 + 1 = 75<\/p>\n\n\n\n

ii. Now, let\u2019s find the sum of 75 numbers
i. e. 1 + 3 + 5 + 7 + \u2026 + 149
\"Maharashtra<\/p>\n\n\n\n

Practice Set 3.4<\/h4>\n\n\n\n

Question 1.
On 1st<\/sup> Jan 2016, Sanika decides to save \u20b9 10, \u20b9 11 on second day, \u20b9 12 on third day. If she decides to save like this, then on 31st Dec 2016 what would be her total saving?
Solution:<\/strong>
\"Maharashtra
\u2234 Sanika\u2019s total saving on 31st<\/sup> December 2016 would be \u20b9 70455.<\/p>\n\n\n\n

Question 2.
A man borrows \u20b9 8000 and agrees to repay with a total interest of \u20b9 1360 in 12 monthly instalments. Each instalment being less than the preceding one by \u20b9 40. Find the amount of the first and last instalment.
Solution:<\/strong>
i. The instalments are in A.P.
Amount repaid in 12 instalments (S12)
= Amount borrowed + total interest
= 8000 + 1360
\u2234 S12<\/sub> = 9360
Number of instalments (n) = 12
Each instalment is less than the preceding one by \u20b9 40.
\u2234 d = -40
\"Maharashtra
\u2234 Amount of the first instalment is \u20b9 1000 and that of the last instalment is \u20b9 560.<\/p>\n\n\n\n

Question 3.
Sachin invested in a national saving certificate scheme. In the first year he invested \u20b9 5000, in the second year \u20b9 7000, in the third year \u20b9 9000 and so on. Find the total amount that he invested in 12 years.
Solution:<\/strong>
i. Amount invested by Sachin in each year are as follows:
5000, 7000, 9000, \u2026
The above sequence is an A.P.
\u2234 a = 5000, d = 7000 \u2013 5000 = 2000, n = 12<\/p>\n\n\n\n

\"Maharashtra
\u2234 The total amount invested by Sachin in 12 years is \u20b9 1,92,000.<\/p>\n\n\n\n

Question 4.
There is an auditorium with 27 rows of seats. There are 20 seats in the first row, 22 seats in the second row, 24 seats in the third row and so on. Find the number of seats in the 15th row and also find how many total seats are there in the auditorium?
Solution:<\/strong>
i. The number of seats arranged row-wise are as follows:
20, 22, 24,
The above sequence is an A.P.
\u2234 a = 20, d = 22 \u2013 20 = 2, n = 27<\/p>\n\n\n\n

ii. tn<\/sub> = a + (n \u2013 1)d
\u2234 t15<\/sub> = 20 + (15 \u2013 1)2
= 20 + 14 \u00d7 2
= 20 + 28
\u2234 t15<\/sub> = 48
\u2234 The number of seats in the 15th<\/sup> row is 48.
\"Maharashtra
\u2234 Total seats in the auditorium are 1242.<\/p>\n\n\n\n

Question 5.
Kargil\u2019s temperature was recorded in a week from Monday to Saturday. All readings were in A.P. The sum of temperatures of Monday and Saturday was 5\u00b0C more than sum of temperatures of Tuesday and Saturday. If temperature of Wednesday was -30\u00b0 Celsius then find the temperature on the other five days.
Solution:<\/strong>
Let the temperatures from Monday to Saturday in A.P. be
a, a + d, a + 2d, a + 3d, a + 4d, a + 5d.
According to the first condition,
(a) + (a + 5d) = (a + d) + (a + 5d) + 5\u00b0
\u2234 d = -5\u00b0
According to the second condition,
a + 2d = -30\u00b0
\u2234 a + 2(-5\u00b0) = -30\u00b0
\u2234 a \u2013 10\u00b0 = -30\u00b0
\u2234 a = -30\u00b0 + 10\u00b0 = -20\u00b0
\u2234 a + d = -20\u00b0 \u2013 5\u00b0 = \u2013 25\u00b0
a + 3d = -20\u00b0 + 3(- 5\u00b0) = -20\u00b0 \u2013 15\u00b0 = -35\u00b0
a + 4d = -20\u00b0 + 4(-5\u00b0) = -20\u00b0 \u2013 20\u00b0 = -40\u00b0
a + 5d = -20\u00b0 + 5(-5\u00b0) = -20\u00b0 \u2013 25\u00b0 = -45\u00b0
\u2234 The temperatures on the other five days are
-20\u00b0C, -25\u00b0 C, -35\u00b0 C, -40\u00b0 C and -45\u00b0 C.<\/p>\n\n\n\n

Question 6.
On the world environment day tree plantation programme was arranged on a land which is triangular in shape. Trees are planted such that in the first row there is one tree, in the second row there are two trees, in the third row three trees and so on. Find the total number of trees in the 25 rows.
Solution:<\/strong>
i. The number of frees planted row-wise are as follows:
1,2,3,\u2026
The above sequence is an A.P.
\u2234 a = 1, d = 2 \u2013 1 = 1,n = 25
\"Maharashtra
\u2234 The total number of trees in 25 rows are 325.<\/p>\n\n\n\n

Download PDF<\/strong><\/h2>\n\n\n\n

Maharashtra Board Solutions for Class 10-Maths (Part 1): Chapter 3- Arithmetic Progression<\/p>\n\n\n\n

Download PDF<\/strong>: Maharashtra Board Solutions for Class 10-Maths (Part 1): Chapter 3- Arithmetic Progression PDF<\/a><\/p>\n\n\n\n

Chapterwise Maharashtra Board Solutions Class 10 Maths (Part 1) :<\/strong><\/h2>\n\n\n\n