{"id":626028,"date":"2023-09-09T07:17:17","date_gmt":"2023-09-09T07:17:17","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=626028"},"modified":"2023-09-09T07:17:22","modified_gmt":"2023-09-09T07:17:22","slug":"kc-sinha-exercise-15-3-mathematics-solution-class-11-chapter-15-anukram-aur-shreni","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-15-3-mathematics-solution-class-11-chapter-15-anukram-aur-shreni\/","title":{"rendered":"KC Sinha: Exercise 15.3 – Mathematics Solution Class 11 Chapter 15 \u0905\u0928\u0941\u0915\u094d\u0930\u092e \u0914\u0930 \u0936\u094d\u0930\u0947\u0923\u0940"},"content":{"rendered":"\n\n\n\n\n

\u0936\u094d\u0930\u0947\u0923\u0940 15+12+9+… \u0915\u0947 \u0915\u093f\u0924\u0928\u0947 \u092a\u0926\u094b \u0915\u094b \u0932\u0947\u0928\u0947 \u0938\u0947 \u0909\u0928\u0915\u093e \u092f\u094b\u0917\u092b\u0932 15 \u092e\u093f\u0932\u0947\u0917\u093e? \u0926\u094b \u0909\u0924\u094d\u0924\u0930\u094b \u0915\u0940 \u0935\u094d\u092f\u093e\u0916\u094d\u092f\u093e \u0915\u0940\u091c\u093f\u090f\u0964<\/strong><\/p>\n\n\n\n

\u0936\u094d\u0930\u0947\u0923\u0940: 15+12+9+…<\/p>\n\n\n\n

-3n2<\/sup>+33n-30=0
-3(n2<\/sup>-11n+10=0)
n2<\/sup>-11n+10=0
n2<\/sup>-10n-n+10=0
n(n-10)-1(n-10)=0<\/p>\n\n\n\n

$\\begin{array}{l|l}n-10=0&n-1=0\\\\n=10&n=1\\end{array}$<\/p>\n\n\n\n

\u0936\u094d\u0930\u0947\u0923\u0940: 15+12+9+…
\u0926\u094d\u0935\u093f\u0924\u0940\u092f \u092a\u0926\u094b \u0938\u0947 \u0936\u0941\u0930\u0942 \u0915\u0930\u0928\u0947 \u092a\u0930 9 \u092a\u0926\u094b \u0915\u093e \u092f\u094b\u0917 =0
n=1 ,10<\/p>\n\n\n\n

Question 4<\/h4>\n\n\n\n

n \u0915\u093e \u092e\u093e\u0928 \u0928\u094d\u092f\u0942\u0928\u0924\u092e \u092e\u093e\u0928 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f \u0924\u093e\u0915\u093f 2+5+8+11+…n \u092a\u0926\u094b \u0924\u0915\u2265200
<\/strong> Sol :
2+5+8+11+…n \u092a\u0926\u094b \u0924\u0915\u2265200<\/p>\n\n\n\n

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

=5-2
=3<\/p>\n\n\n\n

$S_n=\\frac{n}{2}\\left[2a+(n-1)d\\right]$<\/p>\n\n\n\n

$\\frac{n}{2}\\left[2\\times 2+(n-1)3\\right]$\u2265200<\/p>\n\n\n\n

n(3n+1)\u2265400<\/p>\n\n\n\n

3n2<\/sup>+n-400\u22650<\/p>\n\n\n\n

$\\begin{array}{l|l}n\\leq \\frac{-1-\\sqrt{1^2-4\\times 3\\times 400}}{2\\times 3}&n\\geq \\frac{-1-\\sqrt{1^2-4\\times 3\\times (-400)}}{2\\times 3}\\\\n\\leq \\frac{-1-\\sqrt{4801}}{6}&n\\geq \\frac{-1+\\sqrt{4801}}{6}\\\\n\\leq \\frac{-1-69.3}{6}&n\\geq \\frac{-1+69.3}{6}\\\\n\\leq \\frac{-70.3}{6}&n\\geq \\frac{68.3}{6}\\\\n\\leq -11.7&n\\geq 11.3\\end{array}$<\/p>\n\n\n\n

n \u0915\u093e \u092e\u093e\u0928 \u0927\u0928\u093e\u0924\u094d\u092e\u0915 \u0939\u094b\u0917\u093e\u0964
\u2234n=12<\/p>\n\n\n\n

Question 5<\/h4>\n\n\n\n

100 \u0914\u0930 200 \u0915\u0947 \u092c\u0940\u091a \u092e\u0947 \u0938\u094d\u0925\u093f\u0924 \u0938\u092d\u0940 \u0935\u093f\u0937\u092e \u0938\u0902\u0916\u094d\u092f\u093e\u0913 \u0915\u093e \u092f\u094b\u0917\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964
[Find the sum of all odd numbers lying between 100 and 200]
<\/strong> Sol :
100 \u0914\u0930 200 \u0915\u0947 \u092c\u0940\u091a \u092e\u0947 \u0938\u094d\u0925\u093f\u0924 \u0935\u093f\u0937\u092e \u0938\u0902\u0916\u094d\u092f\u093e\u0901\u090f\u0903
101,103,105,….199<\/p>\n\n\n\n

\u092e\u093e\u0928\u093e \u0935\u093f\u0937\u092e \u0938\u0902\u0916\u094d\u092f\u093e\u0913 \u0915\u0940 \u0938\u0902\u0916\u094d\u092f\u093e n \u0939\u0948\u0964
a=101,<\/p>\n\n\n\n

an<\/sub>=199<\/p>\n\n\n\n

a+(n-1)d=199<\/p>\n\n\n\n

101+(n-1)2=1999<\/p>\n\n\n\n

(n-1)2=199-101<\/p>\n\n\n\n

$(n-1)=\\frac{98}{2}$<\/p>\n\n\n\n

n=49+1
n=50<\/p>\n\n\n\n

$S_n=\\frac{n}{2}\\left[a+a^n\\right]$
$=\\frac{50}{2}\\left[101+199\\right]$
=25\u00d7300
=7500<\/p>\n\n\n\n

Question 6<\/h4>\n\n\n\n

\u0924\u0940\u0928 \u0905\u0902\u0915\u094b \u0915\u0940 \u0938\u092d\u0940 \u0938\u0902\u0916\u094d\u092f\u093e\u0913 \u0915\u093e \u092f\u094b\u0917\u092b\u0932 \u091c\u094d\u091e\u0924 \u0915\u0940\u091c\u093f\u090f \u091c\u094b 7 \u0938\u0947 \u0935\u093f\u092d\u093e\u091c\u094d\u092f \u0939\u094b\u0964
[Find the sum of all integers of 3 digits which are divisible by 7]
<\/strong> Sol :
7 \u0938\u0947 \u0935\u093f\u092d\u093e\u091c\u094d\u092f 3 \u0905\u0902\u0915\u0940\u092f \u0938\u0902\u0916\u094d\u092f\u093e
105,112,119….994<\/p>\n\n\n\n

\u092e\u093e\u0928\u093e 7 \u0938\u0947 \u0935\u093f\u092d\u093e\u091c\u094d\u092f 3 \u0905\u0902\u0915\u0940\u092f \u0938\u0902\u0916\u094d\u092f\u093e\u0913 \u0915\u0940 \u0938\u0902\u0916\u094d\u092f\u093e n \u0939\u0948\u0964<\/p>\n\n\n\n

a=105,<\/p>\n\n\n\n

d=a2<\/sub>-a1<\/sub><\/p>\n\n\n\n

=112-105
d=7<\/p>\n\n\n\n

an<\/sub>=994<\/p>\n\n\n\n

a+(n-1)d=994
105+(n-1)7=994
(n-1)7=994-105
(n-1)7=889
$n-1=\\frac{889}{7}$
n=127+1
n=128<\/p>\n\n\n\n

$S_n=\\frac{n}{2}\\left[a+a_n\\right]$<\/p>\n\n\n\n

$S_n=\\frac{128}{2}\\left[105+994\\right]$
=64\u00d71099
=70336<\/p>\n\n\n\n

Question 7<\/h4>\n\n\n\n

1 \u0914\u0930 100 \u0915\u0947 \u092c\u0940\u091a \u092e\u0947 \u0938\u094d\u0925\u093f\u0924 \u0909\u0928 \u0938\u092d\u0940 \u0938\u0902\u0916\u094d\u092f\u093e\u0913 \u0915\u093e \u092f\u094b\u0917\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f \u091c\u094b 2 \u092f\u093e 5 \u0938\u0947 \u0935\u093f\u092d\u093e\u091c\u094d\u092f \u0939\u0948\u0964
<\/strong> Sol :
1 \u0914\u0930 100 \u0915\u0947 \u092c\u0940\u091a \u092e\u0947 2 \u092f\u093e 5 \u0938\u0947 \u0935\u093f\u092d\u093e\u091c\u094d\u092f \u0938\u0902\u0916\u094d\u092f\u093e\u0901\u090f:
(2+4+6+….100)+(5+10+15+….+100)-(10+20+30+…100)<\/p>\n\n\n\n

CASE-I
2 \u0938\u0947 \u0935\u093f\u092d\u093e\u091c\u094d\u092f \u0938\u0902\u0916\u094d\u092f\u093e\u0901\u090f
2+4+6+….+100<\/p>\n\n\n\n

\u092e\u093e\u0928\u093e 2 \u0938\u0947 \u0935\u093f\u092d\u093e\u091c\u094d\u092f \u0909\u092a\u0930 \u0926\u093f\u090f \u0917\u090f \u0905\u0928\u0941\u0915\u094d\u0930\u092e \u092e\u0947 n \u0938\u0902\u0916\u094d\u092f\u093e\u0901\u090f \u0939\u0948\u0964
a=2 ,<\/p>\n\n\n\n

d=a2<\/sub>-a1<\/sub><\/p>\n\n\n\n

=4-2=2<\/p>\n\n\n\n

an<\/sub>=100<\/p>\n\n\n\n

a+(n-1)d=100
2+(n-1)2=100
(n-1)2=98
$n-1=\\frac{98}{2}$
n=50<\/p>\n\n\n\n

$S_n=\\frac{n}{2}\\left[a+a_n\\right]$<\/p>\n\n\n\n

$S_n=\\frac{50}{2}\\left[2+100\\right]$
=25\u00d7102
=2550<\/p>\n\n\n\n

CASE-II
5 \u0938\u0947 \u0935\u093f\u092d\u093e\u091c\u094d\u092f \u0938\u0902\u0916\u094d\u092f\u093e\u0901\u090f:
5+10+15+…100<\/p>\n\n\n\n

\u092e\u093e\u0928\u093e \u0926\u093f\u090f \u0917\u090f \u0905\u0928\u0941\u0915\u094d\u0930\u092e \u092e\u0947 n \u0938\u0902\u0916\u094d\u092f\u093e\u0901\u090f \u0939\u0948\u0964
a=5 ,<\/p>\n\n\n\n

d=10-5
=5

an<\/sub>=100<\/p>\n\n\n\n

a+(n-1)d=100
5+(n-1)5=100
(n-1)5=95
$n-1=\\frac{95}{5}$
n=20<\/p>\n\n\n\n

$S_n=\\frac{20}{2}\\left[5+100\\right]$
=10\u00d7105
=1050<\/p>\n\n\n\n

CASE-III
2 \u0924\u0925\u093e 5 \u0938\u0947 \u0935\u093f\u092d\u093e\u091c\u094d\u092f \u0938\u0902\u0916\u094d\u092f\u093e\u0901\u090f\u0903
10+20+30+….100<\/p>\n\n\n\n

a=10, an<\/sub>=100 , n=10<\/p>\n\n\n\n

$S_n=\\frac{10}{2}\\left[10+100\\right]$<\/p>\n\n\n\n

=5\u00d7110<\/p>\n\n\n\n

=550<\/p>\n\n\n\n

\u22341 \u0938\u0947 100 \u092c\u0940\u091a 2 \u092f\u093e 5 \u0938\u0947 \u0935\u093f\u092d\u093e\u091c\u094d\u092f \u0938\u0902\u0916\u094d\u092f\u093e\u0913 \u0915\u093e \u092f\u094b\u0917
=2550+1050-550
=3050<\/p>\n\n\n\n

Question 8<\/h4>\n\n\n\n

1 \u0914\u0930 200 \u0915\u0947 \u092c\u0940\u091a \u0909\u0928 \u0938\u092d\u0940 \u0938\u0902\u0916\u094d\u092f\u093e\u0913 \u0915\u093e \u091c\u094b\u0921\u093c \u0928\u093f\u0915\u093e\u0932\u093f\u090f \u091c\u093f\u0938\u092e\u0947 3 \u0938\u0947 \u092f\u093e 7 \u0938\u0947 \u092a\u0942\u0930\u093e-\u092a\u0942\u0930\u093e \u092d\u093e\u0917 \u0928\u0939\u0940 \u0932\u0917\u0924\u093e \u0939\u0948 \u0964
<\/strong> Sol :<\/p>\n\n\n\n

Question 10<\/h4>\n\n\n\n

\u0938\u092e\u093e\u0902\u0924\u0930 \u0936\u094d\u0930\u0947\u0922\u093c\u0940 30,27,24,21,18…\u0915\u093e \u092e\u0939\u0924\u094d\u0924\u092e \u092f\u094b\u0917\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964
[Find the maximum sum of the A.P 30,27,24,21,18,…]
<\/strong> Sol :
A.P: 30,27,24,21,18,…<\/p>\n\n\n\n

\u092e\u093e\u0928\u093e A.P \u0915\u093e n \u0935\u093e\u0901 \u092a\u0926 \u0936\u0942\u0928\u094d\u092f \u0938\u0947 \u092c\u0921\u093c\u093e \u0939\u0948\u0964
a=30 ,<\/p>\n\n\n\n

d=27-30=-3<\/p>\n\n\n\n

an<\/sub>>0
a+(n-1)d>0
30+(n-1)(-3)>0
(n-1)(-3)>-30<\/p>\n\n\n\n

$\\frac{(n-1)(-3)}{-3}<\\frac{-30}{-3}$
n<10+1
n<11
\u2234n=10<\/p>\n\n\n\n

$S_n=\\frac{n}{2}\\left[2a+(n-1)d\\right]$<\/p>\n\n\n\n

$S_n=\\frac{10}{2}\\left[2\\times 30+(10-1)(-3)\\right]$
=5[60+9(-3)]
=5[60-27]
=5\u00d733
=165<\/p>\n\n\n\n

Question 11<\/h4>\n\n\n\n

\u0938\u092e\u093e\u0902\u0924\u0930 \u0936\u094d\u0930\u0947\u0922\u093c\u0940 \u0915\u0947 8 \u092a\u0926\u094b \u0915\u093e \u092f\u094b\u0917\u092b\u0932 64 \u0939\u0948 \u0914\u0930 19 \u092a\u0926\u094b \u0915\u093e \u092f\u094b\u0917\u092b\u0932 361 \u0939\u0948 \u0924\u094b n \u092a\u0926\u094b \u0915\u093e \u092f\u094b\u0917\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964
<\/strong> Sol :
$S_n=\\frac{n}{2}\\left[2a+(n-1)d\\right]$<\/p>\n\n\n\n

S8<\/sub>=64 ,S19<\/sub>=64<\/p>\n\n\n\n

$\\frac{8}{2}\\left[2a+(8-1)d\\right]=64$<\/p>\n\n\n\n

$2a+7d=\\frac{64}{4}$<\/p>\n\n\n\n

2a+7d=16..(i)<\/p>\n\n\n\n

$\\frac{19}{2}\\left[2a+(19-1)d\\right]=361$<\/p>\n\n\n\n

$\\frac{19}{2}\\left[2a+18d\\right]=361$<\/p>\n\n\n\n

$\\frac{19}{2}\\times 2 \\left[a+9d\\right]=361$<\/p>\n\n\n\n

$a+9d=\\frac{361}{19}$<\/p>\n\n\n\n

a+9d=19..(i)\u00d72<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u0924\u0925\u093e (ii) \u0938\u0947,<\/p>\n\n\n\n

$\\begin{aligned}2a+7d=&16\\\\2a+18d=&38\\\\\\phantom{2a}-\\phantom{18d}&-\\phantom{38}\\\\ \\hline -11d=-22\\end{aligned}$<\/p>\n\n\n\n

$d=\\frac{22}{11}=2$<\/p>\n\n\n\n

d \u0915\u093e \u092e\u093e\u0928 \u0938\u092e\u0940\u0915\u0930\u0923 (ii) \u092e\u0947 \u0930\u0916\u0928\u0947 \u092a\u0930,
a+9d=19
a+9(2)=19
a+18=19
a=1<\/p>\n\n\n\n

$S_n=\\frac{n}{2}\\left[2\\times 1+(n-1)2\\right]$<\/p>\n\n\n\n

$S_n=\\frac{n}{2}\\left[2+2n-2\\right]$<\/p>\n\n\n\n

$=\\frac{n}{2}\\times 2n=n^2$<\/p>\n\n\n\n

Question 12<\/h4>\n\n\n\n

\u090f\u0915 \u0938\u092e\u093e\u0902\u0924\u0930 \u0936\u094d\u0930\u0947\u0922\u093c\u0940 \u0915\u0947 \u092a\u0939\u0932\u0947 4 \u092a\u0926\u094b \u0915\u093e \u092f\u094b\u0917 \u092a\u0939\u0932\u0947 10 \u092a\u0926\u094b \u0915\u0947 \u092f\u094b\u0917 \u0915\u093e \u0906\u0927\u093e \u0939\u0948 \u0914\u0930 \u092a\u0939\u0932\u0947 (4+p) \u092a\u0926\u094b \u0915\u093e \u092f\u094b\u0917 \u0915\u093e \u092d\u0940 \u0906\u0927\u093e \u0939\u0948, \u091c\u0939\u093e\u0901 p\u22606 \u0924\u094b \u0926\u093f\u0916\u0932\u093e\u092f\u0947 \u0915\u093f p \u0915\u093e \u092e\u093e\u0928 20 \u0939\u0948\u0964
<\/strong> Sol :
$S_n=\\frac{n}{2}\\left[2a+(n-1)d\\right]$<\/p>\n\n\n\n

$S_4=\\frac{1}{2}S_{10}$<\/p>\n\n\n\n

$\\frac{4}{2}\\left[2a+(4-1)d\\right]=\\frac{1}{2}\\times \\frac{10}{2}\\left[2a+(10-1)d\\right]$<\/p>\n\n\n\n

4[2a+3d]=5[2a+9d]
8a+12d=10a+45d
12d-45d=10a-8a
-33d=2a..(i)<\/p>\n\n\n\n

\u0905\u092c,
$S_4=\\frac{1}{2}S_{4+p}$<\/p>\n\n\n\n

$\\frac{4}{2}\\left[2a+(4-1)d\\right]=\\frac{1}{2}\\times \\frac{4+p}{2}\\left[2a+(4+p-1)d\\right]$<\/p>\n\n\n\n

$4[-33d+3d]=\\frac{4+p}{2}\\left[-33d+3d+pd\\right]$<\/p>\n\n\n\n

$4(-30d)=\\frac{4+p}{2}[-30d+pd]$<\/p>\n\n\n\n

$-120d=\\frac{4+p}{2}(-30+p)d$
-240=-120+4p-30p+p2<\/sup>
0=240-120-26p+p2<\/sup>
0=p2<\/sup>-26p+120<\/p>\n\n\n\n

\u092f\u093e p2<\/sup>-26p+120=0
p2<\/sup>-20p-6p+120=0
p(p-20)-6(p-20)=0
(p-20)(p-6)=0<\/p>\n\n\n\n

$\\begin{array}{l|l}p-20=0&p-6=0\\\\p=20&p=6\\end{array}$<\/p>\n\n\n\n

Question 13<\/h4>\n\n\n\n

\u092f\u0926\u093f \u090f\u0915 A.P \u0915\u093e m \u0935\u093e\u0901 \u092a\u0926 $\\frac{1}{n}$ \u0939\u094b \u0914\u0930 n \u0935\u093e\u0901 \u092a\u0926 $\\frac{1}{m}$ \u0939\u094b \u0924\u094b \u0938\u093e\u092c\u093f\u0924 \u0915\u0940\u091c\u093f\u090f \u0915\u093f mn \u092a\u0926\u094b \u0915\u093e \u092f\u094b\u0917\u092b\u0932 $\\frac{mn+1}{2}$ \u0939\u0948\u0964
<\/strong> Sol :
$a_m=\\frac{1}{n}$ , $a_n=\\frac{1}{m}$<\/p>\n\n\n\n

$\\begin{aligned}a+(m-1)d=&\\frac{1}{n}..(i)\\\\a+(n-1)d=&\\frac{1}{m}..(ii)\\\\\\phantom{a}-\\phantom{(n-1)d}&-\\phantom{\\frac{1}{m}}\\\\ \\hline (m-1)d-(n-1)d=&\\frac{1}{n}-\\frac{1}{m}\\end{aligned}$<\/p>\n\n\n\n

$(m-1-n+1)d=\\frac{m-n}{nm}$<\/p>\n\n\n\n

$(m-n)d=\\frac{m-n}{nm}$<\/p>\n\n\n\n

$d=\\frac{1}{mn}$<\/p>\n\n\n\n

d \u0915\u093e \u092e\u093e\u0928 \u0938\u092e\u0940\u0915\u0930\u0923 (i) \u092e\u0947 \u0930\u0916\u0928\u0947 \u092a\u0930,<\/p>\n\n\n\n

$a+(m-1)d=\\frac{1}{n}$<\/p>\n\n\n\n

$a+(m-1)\\frac{1}{mn}=\\frac{1}{n}$<\/p>\n\n\n\n

$a+\\frac{1}{n}-\\frac{1}{mn}=\\frac{1}{n}$<\/p>\n\n\n\n

$a=\\frac{1}{mn}$<\/p>\n\n\n\n

$S_n=\\frac{n}{2}[2a+(n-1)d]$<\/p>\n\n\n\n

$S_{mn}=\\frac{mn}{2}[2\\times \\frac{1}{mn}+(mn-1)\\frac{1}{mn}]$<\/p>\n\n\n\n

$=\\frac{mn}{2}\\left[\\frac{2}{mn}+1-\\frac{1}{mn}\\right]$<\/p>\n\n\n\n

$=\\frac{mn}{2}\\left[\\frac{1}{mn}+1\\right]$<\/p>\n\n\n\n

$=\\frac{mn}{2}\\left[\\frac{1+mn}{mn}\\right]$<\/p>\n\n\n\n

$=\\frac{mn+1}{2}$<\/p>\n\n\n\n

Question 14<\/h4>\n\n\n\n

\u092f\u0926\u093f \u090f\u0915 A.P \u0915\u0947 m \u092a\u0926\u094b \u0915\u093e \u092f\u094b\u0917\u092b\u0932 n \u0939\u094b \u0914\u0930 \u0907\u0938\u0915\u0947 n \u092a\u0926\u094b \u0915\u093e \u092f\u094b\u0917\u092b\u0932 m \u0939\u094b \u0924\u094b \u0926\u093f\u0916\u0932\u093e\u0907\u090f \u0915\u093f (m+n) \u092a\u0926\u094b \u0915\u093e \u092f\u094b\u0917\u092b\u0932 -(m+n) \u0939\u0948\u0964
<\/strong> Sol :
$S_n=\\frac{n}{2}[2a+(n-1)d]$<\/p>\n\n\n\n

Sm<\/sub>=n , Sn<\/sub>=m<\/p>\n\n\n\n

$\\frac{m}{2}[2a+(m-1)d]=n$<\/p>\n\n\n\n

$2a+(m-1)d=\\frac{2n}{m}$..(i)<\/p>\n\n\n\n

Sn<\/sub>=m<\/p>\n\n\n\n

$\\frac{n}{2}[2a+(n-1)d]=m$<\/p>\n\n\n\n

$2a+(n-1)d=\\frac{2m}{n}$..(ii)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u0924\u0925\u093e (ii) \u0938\u0947,<\/p>\n\n\n\n

$\\begin{aligned}2a+(m-1)d=&\\frac{2n}{m}\\\\2a+(n-1)d=&\\frac{2m}{n}\\\\\\phantom{2a}-\\phantom{(n-1)d}&-\\phantom{\\frac{2m}{n}}\\\\ \\hline (m-1)d-(n-1)d=&\\frac{2n}{m}-\\frac{2m}{n} \\end{aligned}$<\/p>\n\n\n\n

$(m-1-n+1)d=\\frac{2n^2-2m^2}{mn}$<\/p>\n\n\n\n

$(m-n)d=\\frac{-2(m^2-n^2)}{mn}$<\/p>\n\n\n\n

$(m-n)d=\\frac{-2(m-n)(m+n)}{mn}$<\/p>\n\n\n\n

$d=\\frac{-2(m+n)}{mn}$<\/p>\n\n\n\n

$d=-2\\left(\\frac{m}{mn}+\\frac{n}{mn}\\right)$<\/p>\n\n\n\n

$d=-\\frac{2}{n}-\\frac{2}{m}$<\/p>\n\n\n\n

d \u0915\u093e \u092e\u093e\u0928 \u0938\u092e\u0940\u0915\u0930\u0923 (i) \u092e\u0947 \u0930\u0916\u0928\u0947 \u092a\u0930,<\/p>\n\n\n\n

$2a+(m-1)d=\\frac{2n}{m}$<\/p>\n\n\n\n

$2a+(m-1)\\left(-\\frac{2}{n}-\\frac{2}{m}\\right)=\\frac{2n}{m}$<\/p>\n\n\n\n

$2a\\left[a+(m-1)\\left[-\\frac{1}{n}-\\frac{1}{m}\\right]\\right]=\\frac{2n}{m}$<\/p>\n\n\n\n

$a-\\frac{m}{n}-1+\\frac{1}{n}+\\frac{1}{m}=\\frac{n}{m}$<\/p>\n\n\n\n

$a=\\frac{n}{m}+\\frac{m}{n}-\\frac{1}{n}-\\frac{1}{m}+1$<\/p>\n\n\n\n

$=\\frac{m+n}{2} \\left[2\\times \\left(\\frac{n}{m}+\\frac{m}{n}-\\frac{1}{n}-\\frac{1}{m}+1\\right)+(m+n-1) \\left(\\frac{-2}{n}-\\frac{2}{m}\\right)\\right]$<\/p>\n\n\n\n

$S_{m+n}=\\frac{m+n}{2}\\left[2\\left(\\frac{n}{m}+\\frac{m}{n}-\\frac{1}{n}-\\frac{1}{m}+1\\right)+(m+n-1) \\left(-\\frac{1}{n}-\\frac{1}{m}\\right)\\right]$<\/p>\n\n\n\n

$=(m+n) \\left[\\frac{n}{m}+\\frac{m}{n}-\\frac{1}{n}-\\frac{1}{m}+1-\\frac{m}{n}-1-1-\\frac{n}{m}+\\frac{1}{n}+\\frac{1}{m}\\right]$<\/p>\n\n\n\n

=(m+n)(-1)=-(m+n)<\/p>\n\n\n\n

Question 15<\/h4>\n\n\n\n

\u0915\u093f\u0938\u0940 A.P \u0915\u0947 n, 2n \u0924\u0925\u093e 3n \u092a\u0926\u094b \u0915\u093e \u091c\u094b\u0921\u093c \u0915\u094d\u0930\u092e\u0936\u0903 S1<\/sub>,S2 <\/sub>\u0914\u0930 S3 <\/sub>\u0939\u094b \u0924\u094b \u0926\u093f\u0916\u093e\u0932\u093e\u0907\u090f \u0915\u093f S3 <\/sub>=3(S2<\/sub>-S1<\/sub>)
<\/strong> Sol :
Sn<\/sub>=S1 <\/sub>,S2n<\/sub>=S2 <\/sub>,S3n<\/sub>=S3<\/sub><\/p>\n\n\n\n

R.H.S
3(S2<\/sub>-S1<\/sub>)<\/p>\n\n\n\n

=3[S2n<\/sub>-Sn<\/sub>]<\/p>\n\n\n\n

$=3\\left[\\frac{2n}{2}\\left\\{2a+(2n-1)d\\right\\}-\\frac{n}{2}\\left\\{2a+(n-1)d\\right\\}\\right]$<\/p>\n\n\n\n

$=\\frac{3n}{2}\\left[2{2a+(2n-1)d}\\right]-{2a+(n-1)d}$<\/p>\n\n\n\n

$=\\frac{3n}{2}\\left[4a+(4n-2)d-2a-(n-1)d\\right]$<\/p>\n\n\n\n

$=\\frac{3n}{2}\\left[2a+(4n-2-n+1)d\\right]$<\/p>\n\n\n\n

$=\\frac{3n}{2}\\left[2a+(3n-1)d\\right]=S_{3n}=S_3$<\/p>\n\n\n\n

Question 16<\/h4>\n\n\n\n

\u092f\u0926\u093f \u0915\u093f\u0938\u0940 A.P \u0915\u0947 (2n+1) \u092a\u0926\u094b \u0915\u093e \u091c\u094b\u0921\u093c S \u0939\u094b \u0914\u0930 \u092a\u094d\u0930\u0925\u092e \u092a\u0926 \u0938\u0947 \u092a\u094d\u0930\u093e\u0930\u092e\u094d\u092e \u0915\u0930 \u0915\u094d\u0930\u092e\u093e\u0928\u094d\u0924\u0930 (alternate) \u092a\u0926\u094b \u0915\u093e \u091c\u094b\u0921\u093c S1 <\/sub>\u0939\u094b \u0924\u094b \u0926\u093f\u0916\u0932\u093e\u0907\u090f \u0915\u093f $\\frac{S}{S_1}=\\frac{2n+1}{n+1}$
<\/strong> Sol :
A.P: a,a+d,a+2d,a+3d….a2n+1<\/sub><\/p>\n\n\n\n

S=S2n+1<\/sub>$=\\frac{2n+1}{2}\\left[2a+(2n+1-1)d\\right]$<\/p>\n\n\n\n

$S=\\frac{2n+1}{2}[2a+2nd]$..(i)<\/p>\n\n\n\n

$S_1=\\frac{n+1}{2}[2a+(n+1-1)2d]$<\/p>\n\n\n\n

$S_1=\\frac{n+1}{2}[2a+2nd]$..(ii)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u092e\u0947 (ii) \u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930,<\/p>\n\n\n\n

$\\frac{S}{S_1}=\\dfrac{\\frac{2n+1}{2}[2a+2nd]}{\\frac{n+1}{2}[2a+2nd]}$<\/p>\n\n\n\n

$=\\frac{2n+1}{n+1}$<\/p>\n\n\n\n

Question 17<\/h4>\n\n\n\n

\u092f\u0926\u093f \u0915\u093f\u0938\u0940 A.P \u0915\u0947 \u092a\u094d\u0930\u0925\u092e \u0924\u0943\u0924\u0940\u092f \u0924\u0925\u093e n \u0935\u0947 \u092a\u0926 \u0915\u094d\u0930\u092e\u0936\u0903 a,b,c \u0939\u094b \u0924\u094b \u0938\u093e\u092c\u093f\u0924 \u0915\u0940\u091c\u093f\u090f \u0915\u093f n \u092a\u0926\u094b \u0915\u093e \u092f\u094b\u0917 \u0939\u0948\u0903 $\\frac{c+a}{2}+\\frac{c^2-a^2}{b-a}$
<\/strong> Sol :
\u092a\u094d\u0930\u0925\u092e \u092a\u0926=a , a3<\/sub>=b, an<\/sub>=c<\/p>\n\n\n\n

$\\begin{array}{l|l}a_3=b&a_n=c\\\\a+2d=b&a+(n-1)d=c\\\\2d=b-a&a+(n-1)\\left(\\frac{b-a}{2}\\right)=c\\\\d=\\frac{b-a}{2}&(n-1)\\left(\\frac{b-a}{2}\\right)=c-a\\\\&n-1=\\frac{2(c-a)}{b-a}\\\\&n=\\frac{2(c-a)}{b-a}+1\\end{array}$<\/p>\n\n\n\n

$S_n=\\frac{n}{2}\\left[a+a_n\\right]$<\/p>\n\n\n\n

$=\\dfrac{\\left\\{\\frac{2(c-a)}{b-a}+1\\right\\}[a+c]}{2}$<\/p>\n\n\n\n

$=\\left[\\frac{c-a}{b-a}+\\frac{1}{2}\\right](c+a)$<\/p>\n\n\n\n

$S_n=\\frac{c^2-a^2}{b-a}+\\frac{c+a}{2}$<\/p>\n\n\n\n

$=\\frac{c+a}{2}+\\frac{c^2-a^2}{b-a}$<\/p>\n\n\n\n

Question 18<\/h4>\n\n\n\n

\u092f\u0926\u093f \u0915\u093f\u0938\u0940 A.P \u0915\u093e p \u0935\u093e\u0901 \u092a\u0926 x \u0939\u0948 \u0914\u0930 q \u0935\u093e\u0901 \u092a\u0926 y \u0939\u0948 \u0924\u094b \u0926\u093f\u0916\u0932\u093e\u0907\u090f \u0915\u093f (p+q) \u092a\u0926\u094b \u0915\u093e \u091c\u094b\u0921\u093c \u0939\u0948:
[If the pth term of an A.P is x and q is term is y. Show that the sum of (p+q)terms is
$\\frac{p+q}{2}\\left\\{x+y+\\frac{x-y}{p-q}\\right\\}$]
<\/strong> Sol :
ap<\/sub>=x , aq<\/sub>=y<\/p>\n\n\n\n

$\\begin{aligned}a+(p-1)d=&x..(i)\\\\a+(q-1)d=&y..(ii)\\\\-\\phantom{a}-\\phantom{(q-1)d}&-\\phantom{y} \\\\\\hline (p-1)d-(q-1)d=&x-y\\end{aligned}$<\/p>\n\n\n\n

(p-1-q+1)d=x-y<\/p>\n\n\n\n

$d=\\frac{x-y}{p-q}$<\/p>\n\n\n\n

d \u0915\u093e \u092e\u093e\u0928 \u0938\u092e\u0940\u0915\u0930\u0923 (i) \u092e\u0947 \u0930\u0916\u0928\u0947 \u092a\u0930,<\/p>\n\n\n\n

a+(p-1)d=x<\/p>\n\n\n\n

$a+(p-1)\\left(\\frac{x-y}{p-q}\\right)=x$<\/p>\n\n\n\n

$a=x-\\frac{(p-1)(x-y)}{p-q}$<\/p>\n\n\n\n

$S_n=\\frac{n}{2}\\left[2a+(n-1)d\\right]$<\/p>\n\n\n\n

$S_{p+q}=\\frac{p+q}{2}\\left[2\\left\\{x-\\frac{(p-1)(x-y)}{p-q}\\right\\}+(p+q-1)\\frac{(x-y)}{p-q}\\right]$<\/p>\n\n\n\n

$=\\frac{p+q}{2}\\left[2x-\\frac{2(p-1)(x-y)}{p-q}+\\frac{(p-1)(x-y)}{p-q}+\\frac{q(x-y)}{p-q}\\right]$<\/p>\n\n\n\n

$=\\frac{p+q}{2}\\left[2x-\\frac{(p-1)(x-y)}{p-q}+\\frac{q(x-y)}{p-q}\\right]$<\/p>\n\n\n\n

$S-{p+q}=\\frac{p+q}{2}\\left[2x-(p-1-q)\\left(\\frac{x-y}{p-q}\\right)\\right]$<\/p>\n\n\n\n

$=\\frac{p+q}{2}\\left[2x-{(p-q)-1}\\left(\\frac{x-y}{p-q}\\right)\\right]$<\/p>\n\n\n\n

$=\\frac{p-q}{2}\\left[2x-(x-y)+\\frac{x-y}{p-q}\\right]$<\/p>\n\n\n\n

$=\\frac{p+q}{2}\\left[2x-x+y+\\frac{x-y}{p-q}\\right]$<\/p>\n\n\n\n

$S_{p+q}=\\frac{p+q}{2}\\left[x+y+\\frac{x-y}{p-q}\\right]$<\/p>\n\n\n\n

Question 19<\/h4>\n\n\n\n

(i) \u0926\u094b A.P \u0915\u0947 n \u092a\u0926\u094b \u0915\u093e \u091c\u094b\u0921\u093c (3n+8):(7n+15) \u0915\u0947 \u0905\u0928\u0941\u092a\u093e\u0924 \u092e\u0947 \u0939\u0948 \u0924\u094b \u0909\u0928\u0915\u0947 12 \u0935\u0947 \u092a\u0926\u094b \u0915\u093e \u0905\u0928\u0941\u092a\u093e\u0924 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f \u0964
<\/strong> Sol :
$\\frac{S_n}{S_n}=\\frac{3n+8}{7n+15}$<\/p>\n\n\n\n

$\\dfrac{\\frac{n}{2}[2a+(n-1)d]}{\\frac{n}{2}[2A+(n-1)D]}=\\frac{3n+8}{7n+15}$<\/p>\n\n\n\n

n=23 \u092a\u0930,<\/p>\n\n\n\n

$\\frac{2a+(23-1)d}{2A+(23-1)D}=\\frac{3(23)+8}{7(23)+15}$<\/p>\n\n\n\n

$\\frac{2a+22d}{2A+22D}=\\frac{69+8}{161+15}$<\/p>\n\n\n\n

$\\frac{2(a+11d)}{2(A+11D)}=\\frac{77}{176}$<\/p>\n\n\n\n

$\\frac{a_{12}}{A_{12}}=\\frac{7}{16}$<\/p>\n\n\n\n

=7:16<\/p>\n\n\n\n

Question 20<\/h4>\n\n\n\n

\u092f\u0926\u093f \u0915\u093f\u0938\u0940 A.P \u0915\u0947 m \u092a\u0926\u094b \u0915\u093e \u092f\u094b\u0917\u092b\u0932 \u0924\u0925\u093e n \u092a\u0926\u094b \u0915\u0947 \u092f\u094b\u0917\u092b\u0932 \u0915\u093e \u0905\u0928\u0941\u092a\u093e\u0924 m2<\/sup>:n2 <\/sup>\u0939\u094b \u0924\u094b \u0938\u093e\u092c\u093f\u0924 \u0915\u0940\u091c\u093f\u090f \u0915\u093f \u0907\u0938\u0915\u0947 m \u0935\u093e\u0901 \u092a\u0926\u094b \u0915\u093e \u0905\u0928\u0941\u092a\u093e\u0924 (2m-1):(2n-1) \u0939\u0948\u0964
<\/strong> Sol :
$\\frac{S_m}{S_n}=\\frac{m^2}{n^2}$<\/p>\n\n\n\n

$\\dfrac{\\frac{m}{2}[2a+(m-1)d]}{\\frac{n}{2}[2a+(n-1)d]}=\\frac{m^2}{n^2}$<\/p>\n\n\n\n

$\\frac{2a+(m-1)d}{2a+(n-1)d}=\\frac{m^2}{n^2}\\times \\frac{n}{m}$<\/p>\n\n\n\n

m \u0915\u0947 \u091c\u0917\u0939 2m-1 \u0924\u0925\u093e n \u0915\u0947 \u091c\u0917\u0939 2n-1 \u0930\u0916\u0928\u0947 \u092a\u0930,<\/p>\n\n\n\n

$\\frac{2a+(2m-1-1)d}{2a+(2n-1-1)d}=\\frac{2m-1}{2n-1}$<\/p>\n\n\n\n

$\\frac{2a+(2m-2)d}{2a+(2n-2)d}=\\frac{2m-1}{2n-1}$<\/p>\n\n\n\n

$\\frac{2[a+(m-1)d]}{2[a+(n-1)d]}=\\frac{2m-1}{2n-1}$<\/p>\n\n\n\n

$\\frac{a_m}{a_n}=(2m-1):(2n-1)$<\/p>\n\n\n\n

Question 21<\/h4>\n\n\n\n

\u090f\u0915 \u0935\u094d\u092f\u0915\u094d\u0924\u093f 700 \u092a\u094d\u0930\u0924\u093f \u092e\u093e\u0939 \u0935\u0947\u0924\u0928 \u092a\u0930 \u0928\u093f\u092f\u0941\u0915\u094d\u0924 \u0915\u093f\u092f\u093e \u0917\u092f\u093e\u0964 \u0928\u093f\u092f\u0941\u0915\u094d\u0924\u093f \u0915\u0940 \u0936\u0930\u094d\u0924\u0947 \u0925\u0940 \u0915\u093f \u0926\u0942\u0938\u0930\u0947 \u0935\u0930\u094d\u0937 \u0909\u0938\u0947 740 \u092a\u094d\u0930\u0924\u093f \u092e\u093e\u0939, \u0924\u0940\u0938\u0930\u0947 \u0935\u0930\u094d\u0937 780 \u092a\u094d\u0930\u0924\u093f \u092e\u093e\u0939 \u0914\u0930 \u0907\u0938\u0940 \u0924\u0930\u0939 \u0935\u0947\u0924\u0928 \u0935\u0943\u0926\u094d\u0927\u093f \u0939\u094b\u0917\u0940\u0964 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f \u0915\u093f 10 \u0935\u0930\u094d\u0937 \u0915\u0947 \u0905\u0928\u094d\u0924 \u0924\u0915 \u0909\u0938\u0947 \u0915\u0941\u0932 \u0915\u093f\u0924\u0928\u093e \u0935\u0947\u0924\u0928 \u092e\u093f\u0932\u0917\u093e\u0964
<\/strong> Sol :
\u092a\u094d\u0930\u0925\u092e 10 \u0935\u0930\u094d\u0937\u094b \u0915\u093e \u0935\u0947\u0924\u0928\u0903
8400,8880,9360…10 \u0935\u0930\u094d\u0937\u094b \u0924\u0915<\/p>\n\n\n\n

a=8400 ,<\/p>\n\n\n\n

d=a2<\/sub>-a1<\/sub>
=8880-8400
d=480<\/p>\n\n\n\n

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

10 \u0935\u0930\u094d\u0937\u094b \u0924\u0915 \u0915\u0941\u0932 \u0935\u0947\u0924\u0928 $S_n=\\frac{n}{2}[2a+(n-1)d]$<\/p>\n\n\n\n

$=\\frac{10}{2}[2\\times 8400 +(10-1)480]$<\/p>\n\n\n\n

=5[16800+4320]
=5\u00d721120
=105600<\/p>\n\n\n\n

Question 22<\/h4>\n\n\n\n

\u0915\u093f\u0938\u0940 \u092c\u0939\u0941\u092d\u0941\u091c \u0915\u0947 \u0905\u0928\u094d\u0924\u0903 \u0915\u094b\u0923 A.P \u092e\u0947 \u0939\u0948\u0964 \u092f\u0926\u093f \u0928\u094d\u092f\u0942\u0928\u0924\u092e \u0915\u094b\u0923 75\u00b0 \u0939\u094b \u0924\u0925\u093e \u092a\u0926\u093e\u0902\u0924\u0930 10\u00b0 \u0939\u094b \u0924\u094b \u092d\u0941\u091c\u093e\u0913 \u0915\u0940 \u0938\u0902\u0916\u094d\u092f\u093e \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964
<\/strong> Sol :
\u092e\u093e\u0928\u093e \u092c\u0939\u0941\u092d\u0941\u091c \u092e\u0947 n \u092d\u0941\u091c\u093e\u090f\u0901 \u0939\u0948\u0964
\u0928\u094d\u092f\u0942\u0928\u0924\u092e \u0915\u094b\u0923 , a=75\u00b0 ,d=10\u00b0<\/p>\n\n\n\n

$S_n=\\frac{n}{2}[2a+(n-1)d]$<\/p>\n\n\n\n

$=\\frac{n}{2}[2\\times 75 +(n-1)10]$<\/p>\n\n\n\n

$=\\frac{n}{2}[150 +10n-10]$<\/p>\n\n\n\n

$=\\frac{n}{2}[10n+140]$<\/p>\n\n\n\n

Sn<\/sub>=5n2<\/sup>+70n<\/p>\n\n\n\n

\u2235n \u092d\u0941\u091c\u093e\u0913 \u0935\u093e\u0932\u0947 \u092c\u0939\u0941\u092d\u0941\u091c \u0915\u0947 \u0905\u0902\u0924\u0903 \u0915\u094b\u0923\u094b \u0915\u093e \u092f\u094b\u0917
=(n-2)\u00d7180\u00b0<\/p>\n\n\n\n

\u22345n2<\/sup>+70n=(n-2)\u00d7180\u00b0
5(n2<\/sup>+14n)=(n-2)\u00d7180\u00b0
n2<\/sup>+14n$=\\frac{(n-2)\\times 180}{5}$
n2<\/sup>+14n=36n-72
n2<\/sup>+14n-36n+72=0
n2<\/sup>+22n+72=0
n2<\/sup>-18n-4n+72=0
n(n-18)-4(n-18)=0
(n-18)(n-4)=0<\/p>\n\n\n\n

$\\begin{array}{l|l}n-18=0&n-4=0\\\\n=18&n=4\\end{array}$<\/p>\n\n\n\n

a18<\/sub>=a+17d
=75+17(10)
=75+170=245<\/p>\n\n\n\n

\u092c\u0939\u0941\u092d\u0941\u091c \u0915\u093e \u092a\u094d\u0930\u0924\u094d\u092f\u0947\u0915 \u0905\u0902\u0924\u093e \u0915\u094b\u0923 180\u00b0 \u0938\u0947 \u091b\u094b\u091f\u093e \u0939\u094b\u0928\u093e \u091a\u093e\u0939\u093f\u090f\u0964
\u2234\u092d\u0941\u091c\u093e\u0913 \u0915\u0940 \u0938\u0902\u0916\u094d\u092f\u093e=4<\/p>\n\n\n\n

Question 23<\/h4>\n\n\n\n

10 \u092a\u094d\u0930\u0936\u094d\u0928\u094b \u0915\u0940 \u091c\u093e\u0901\u0924 \u092a\u0930\u093f\u0915\u094d\u0937\u093e \u092e\u0947 \u0915\u093f\u0938\u0940 \u0935\u093f\u0926\u094d\u092f\u093e\u0930\u094d\u0925\u0940 \u0915\u094b \u0915\u0939\u093e \u091c\u093e\u0924\u093e \u0939\u0948 \u0915\u093f \u092c\u093e\u0926 \u0915\u0947 \u092a\u094d\u0930\u0936\u094d\u0928 \u0915\u093e \u0905\u0902\u0915 \u0920\u0940\u0915 \u092a\u0939\u0932\u0947 \u0915\u0947 \u092a\u094d\u0930\u0936\u094d\u0928 \u0915\u0947 \u0905\u0902\u0915 \u0938\u0947 2 \u0905\u0927\u093f\u0915 \u0939\u0948\u0964 \u092f\u0926\u093f \u091c\u093e\u0901\u091a \u092a\u0930\u0940\u0915\u094d\u0937\u093e \u0915\u0947 \u092a\u094d\u0930\u0936\u094d\u0928 \u0915\u093e \u0924\u0940\u0938\u0930\u093e \u092a\u094d\u0930\u0936\u094d\u0928 5 \u0905\u0902\u0915 \u0915\u093e \u0939\u094b \u0924\u094b \u0935\u0939 \u0935\u093f\u0926\u094d\u092f\u093e\u0930\u094d\u0925\u0940 8 \u092a\u094d\u0930\u0936\u094d\u0928\u094b \u0915\u094b \u0939\u0932 \u0915\u0930 \u0905\u0927\u093f\u0915 \u0938\u0947 \u0905\u0927\u093f\u0915 \u0915\u093f\u0924\u0928\u093e \u0905\u0902\u0915 \u092a\u094d\u0930\u093e\u092a\u094d\u0924 \u0915\u0930 \u0938\u0915\u0947\u0917\u093e\u0964
<\/strong> Sol :
\u0924\u0940\u0938\u0930\u0947 \u092a\u094d\u0930\u0936\u094d\u0928 \u0938\u0947 \u0936\u0941\u0930\u0942 \u0915\u0930\u0928\u0947 \u092a\u0930 \u092a\u094d\u0930\u0936\u094d\u0928\u094b \u0915\u0947 \u092a\u094d\u0930\u093e\u0923\u094d\u0924\u093e\u0902\u0915
5+7+9+….8 \u092a\u094d\u0930\u0936\u094d\u0928\u094b \u0924\u0915
a=5 ,<\/p>\n\n\n\n

d=a2<\/sub>-a1<\/sub>
=7-5=2<\/p>\n\n\n\n

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

\n
KC Sinha Solutions<\/a><\/div>\n\n\n\n
KC Sinha Class 11 Solutions<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

\u0936\u094d\u0930\u0947\u0923\u0940 15+12+9+… \u0915\u0947 \u0915\u093f\u0924\u0928\u0947 \u092a\u0926\u094b \u0915\u094b \u0932\u0947\u0928\u0947 \u0938\u0947 \u0909\u0928\u0915\u093e \u092f\u094b\u0917\u092b\u0932 15 \u092e\u093f\u0932\u0947\u0917\u093e? \u0926\u094b \u0909\u0924\u094d\u0924\u0930\u094b \u0915\u0940 \u0935\u094d\u092f\u093e\u0916\u094d\u092f\u093e \u0915\u0940\u091c\u093f\u090f\u0964 \u0936\u094d\u0930\u0947\u0923\u0940: 15+12+9+… -3n2+33n-30=0-3(n2-11n+10=0)n2-11n+10=0n2-10n-n+10=0n(n-10)-1(n-10)=0 $\\begin{array}{l|l}n-10=0&n-1=0\\\\n=10&n=1\\end{array}$ \u0936\u094d\u0930\u0947\u0923\u0940: 15+12+9+…\u0926\u094d\u0935\u093f\u0924\u0940\u092f \u092a\u0926\u094b \u0938\u0947 \u0936\u0941\u0930\u0942 \u0915\u0930\u0928\u0947 \u092a\u0930 9 \u092a\u0926\u094b \u0915\u093e \u092f\u094b\u0917 =0n=1 ,10 Question 4 n \u0915\u093e \u092e\u093e\u0928 \u0928\u094d\u092f\u0942\u0928\u0924\u092e \u092e\u093e\u0928 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f \u0924\u093e\u0915\u093f 2+5+8+11+…n \u092a\u0926\u094b \u0924\u0915\u2265200 Sol :2+5+8+11+…n \u092a\u0926\u094b \u0924\u0915\u2265200 a=2 , =5-2=3 $S_n=\\frac{n}{2}\\left[2a+(n-1)d\\right]$ […]<\/p>\n","protected":false},"author":302,"featured_media":626030,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[919],"tags":[],"boards":[],"class_list":["post-626028","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-class-11","entry"],"yoast_head":"\nKC Sinha: Exercise 15.3 - Mathematics Solution Class 11 Chapter 15 \u0905\u0928\u0941\u0915\u094d\u0930\u092e \u0914\u0930 \u0936\u094d\u0930\u0947\u0923\u0940 - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"\u0936\u094d\u0930\u0947\u0923\u0940 15+12+9+... \u0915\u0947 \u0915\u093f\u0924\u0928\u0947 \u092a\u0926\u094b \u0915\u094b \u0932\u0947\u0928\u0947 \u0938\u0947 \u0909\u0928\u0915\u093e \u092f\u094b\u0917\u092b\u0932 15 \u092e\u093f\u0932\u0947\u0917\u093e? \u0926\u094b \u0909\u0924\u094d\u0924\u0930\u094b \u0915\u0940 \u0935\u094d\u092f\u093e\u0916\u094d\u092f\u093e \u0915\u0940\u091c\u093f\u090f\u0964 \u0936\u094d\u0930\u0947\u0923\u0940: 15+12+9+...\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-15-3-mathematics-solution-class-11-chapter-15-anukram-aur-shreni\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"KC Sinha: Exercise 15.3 - Mathematics Solution Class 11 Chapter 15 \u0905\u0928\u0941\u0915\u094d\u0930\u092e \u0914\u0930 \u0936\u094d\u0930\u0947\u0923\u0940\" \/>\n<meta property=\"og:description\" content=\"\u0936\u094d\u0930\u0947\u0923\u0940 15+12+9+... \u0915\u0947 \u0915\u093f\u0924\u0928\u0947 \u092a\u0926\u094b \u0915\u094b \u0932\u0947\u0928\u0947 \u0938\u0947 \u0909\u0928\u0915\u093e \u092f\u094b\u0917\u092b\u0932 15 \u092e\u093f\u0932\u0947\u0917\u093e? \u0926\u094b \u0909\u0924\u094d\u0924\u0930\u094b \u0915\u0940 \u0935\u094d\u092f\u093e\u0916\u094d\u092f\u093e \u0915\u0940\u091c\u093f\u090f\u0964 \u0936\u094d\u0930\u0947\u0923\u0940: 15+12+9+...\" \/>\n<meta property=\"og:url\" content=\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-15-3-mathematics-solution-class-11-chapter-15-anukram-aur-shreni\/\" \/>\n<meta property=\"og:site_name\" content=\"IndCareer Schools\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/indcareer\" \/>\n<meta property=\"article:published_time\" content=\"2023-09-09T07:17:17+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2023-09-09T07:17:22+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-16-1-scaled.jpg\" \/>\n\t<meta property=\"og:image:width\" content=\"1600\" \/>\n\t<meta property=\"og:image:height\" content=\"901\" \/>\n\t<meta property=\"og:image:type\" content=\"image\/jpeg\" \/>\n<meta name=\"author\" content=\"Pooja\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@indcareer\" \/>\n<meta name=\"twitter:site\" content=\"@indcareer\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Pooja\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"6 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-15-3-mathematics-solution-class-11-chapter-15-anukram-aur-shreni\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-15-3-mathematics-solution-class-11-chapter-15-anukram-aur-shreni\/\"},\"author\":{\"name\":\"Pooja\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/#\/schema\/person\/d6945cf059726f162259ba738092301e\"},\"headline\":\"KC Sinha: Exercise 15.3 – Mathematics Solution Class 11 Chapter 15 \u0905\u0928\u0941\u0915\u094d\u0930\u092e \u0914\u0930 \u0936\u094d\u0930\u0947\u0923\u0940\",\"datePublished\":\"2023-09-09T07:17:17+00:00\",\"dateModified\":\"2023-09-09T07:17:22+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-15-3-mathematics-solution-class-11-chapter-15-anukram-aur-shreni\/\"},\"wordCount\":1777,\"publisher\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/#organization\"},\"image\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-15-3-mathematics-solution-class-11-chapter-15-anukram-aur-shreni\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-16-1-scaled.jpg\",\"articleSection\":[\"class 11\"],\"inLanguage\":\"en-US\"},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-15-3-mathematics-solution-class-11-chapter-15-anukram-aur-shreni\/\",\"url\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-15-3-mathematics-solution-class-11-chapter-15-anukram-aur-shreni\/\",\"name\":\"KC Sinha: Exercise 15.3 - Mathematics Solution Class 11 Chapter 15 \u0905\u0928\u0941\u0915\u094d\u0930\u092e \u0914\u0930 \u0936\u094d\u0930\u0947\u0923\u0940 - IndCareer Schools\",\"isPartOf\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/#website\"},\"primaryImageOfPage\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-15-3-mathematics-solution-class-11-chapter-15-anukram-aur-shreni\/#primaryimage\"},\"image\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-15-3-mathematics-solution-class-11-chapter-15-anukram-aur-shreni\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-16-1-scaled.jpg\",\"datePublished\":\"2023-09-09T07:17:17+00:00\",\"dateModified\":\"2023-09-09T07:17:22+00:00\",\"description\":\"\u0936\u094d\u0930\u0947\u0923\u0940 15+12+9+... \u0915\u0947 \u0915\u093f\u0924\u0928\u0947 \u092a\u0926\u094b \u0915\u094b \u0932\u0947\u0928\u0947 \u0938\u0947 \u0909\u0928\u0915\u093e \u092f\u094b\u0917\u092b\u0932 15 \u092e\u093f\u0932\u0947\u0917\u093e? \u0926\u094b \u0909\u0924\u094d\u0924\u0930\u094b \u0915\u0940 \u0935\u094d\u092f\u093e\u0916\u094d\u092f\u093e \u0915\u0940\u091c\u093f\u090f\u0964 \u0936\u094d\u0930\u0947\u0923\u0940: 15+12+9+...\",\"breadcrumb\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-15-3-mathematics-solution-class-11-chapter-15-anukram-aur-shreni\/#breadcrumb\"},\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-15-3-mathematics-solution-class-11-chapter-15-anukram-aur-shreni\/\"]}]},{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-15-3-mathematics-solution-class-11-chapter-15-anukram-aur-shreni\/#primaryimage\",\"url\":\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-16-1-scaled.jpg\",\"contentUrl\":\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-16-1-scaled.jpg\",\"width\":1600,\"height\":901,\"caption\":\"KC Sinha: Exercise 15.10 - Mathematics Solution Class 11 Chapter 15 \u0905\u0928\u0941\u0915\u094d\u0930\u092e \u0914\u0930 \u0936\u094d\u0930\u0947\u0923\u0940\"},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-15-3-mathematics-solution-class-11-chapter-15-anukram-aur-shreni\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https:\/\/www.indcareer.com\/schools\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"class 11\",\"item\":\"https:\/\/www.indcareer.com\/schools\/class-11\/\"},{\"@type\":\"ListItem\",\"position\":3,\"name\":\"KC Sinha: Exercise 15.3 – Mathematics Solution Class 11 Chapter 15 \u0905\u0928\u0941\u0915\u094d\u0930\u092e \u0914\u0930 \u0936\u094d\u0930\u0947\u0923\u0940\"}]},{\"@type\":\"WebSite\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/#website\",\"url\":\"https:\/\/www.indcareer.com\/schools\/\",\"name\":\"IndCareer Schools\",\"description\":\"School Admissions & Notices\",\"publisher\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/#organization\"},\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/www.indcareer.com\/schools\/?s={search_term_string}\"},\"query-input\":{\"@type\":\"PropertyValueSpecification\",\"valueRequired\":true,\"valueName\":\"search_term_string\"}}],\"inLanguage\":\"en-US\"},{\"@type\":\"Organization\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/#organization\",\"name\":\"IndCareer\",\"url\":\"https:\/\/www.indcareer.com\/schools\/\",\"logo\":{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/#\/schema\/logo\/image\/\",\"url\":\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/06\/indcareer-logo2.png\",\"contentUrl\":\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/06\/indcareer-logo2.png\",\"width\":512,\"height\":250,\"caption\":\"IndCareer\"},\"image\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/#\/schema\/logo\/image\/\"},\"sameAs\":[\"https:\/\/www.facebook.com\/indcareer\",\"https:\/\/x.com\/indcareer\",\"https:\/\/www.youtube.com\/channel\/UC1liU3RZoBRuu8YcAuZMsOQ\"],\"email\":\"info@ebharat.in\",\"legalName\":\"IndCareer\",\"numberOfEmployees\":{\"@type\":\"QuantitativeValue\",\"minValue\":\"1\",\"maxValue\":\"10\"}},{\"@type\":\"Person\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/#\/schema\/person\/d6945cf059726f162259ba738092301e\",\"name\":\"Pooja\",\"image\":{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/#\/schema\/person\/image\/\",\"url\":\"https:\/\/secure.gravatar.com\/avatar\/350f7cfdfb6a23bcab67b56b5e77549db2a13b5d23e63175ac5bd07b5d44b720?s=96&d=mm&r=g\",\"contentUrl\":\"https:\/\/secure.gravatar.com\/avatar\/350f7cfdfb6a23bcab67b56b5e77549db2a13b5d23e63175ac5bd07b5d44b720?s=96&d=mm&r=g\",\"caption\":\"Pooja\"}}]}<\/script>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"KC Sinha: Exercise 15.3 - Mathematics Solution Class 11 Chapter 15 \u0905\u0928\u0941\u0915\u094d\u0930\u092e \u0914\u0930 \u0936\u094d\u0930\u0947\u0923\u0940 - IndCareer Schools","description":"\u0936\u094d\u0930\u0947\u0923\u0940 15+12+9+... \u0915\u0947 \u0915\u093f\u0924\u0928\u0947 \u092a\u0926\u094b \u0915\u094b \u0932\u0947\u0928\u0947 \u0938\u0947 \u0909\u0928\u0915\u093e \u092f\u094b\u0917\u092b\u0932 15 \u092e\u093f\u0932\u0947\u0917\u093e? \u0926\u094b \u0909\u0924\u094d\u0924\u0930\u094b \u0915\u0940 \u0935\u094d\u092f\u093e\u0916\u094d\u092f\u093e \u0915\u0940\u091c\u093f\u090f\u0964 \u0936\u094d\u0930\u0947\u0923\u0940: 15+12+9+...","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-15-3-mathematics-solution-class-11-chapter-15-anukram-aur-shreni\/","og_locale":"en_US","og_type":"article","og_title":"KC Sinha: Exercise 15.3 - Mathematics Solution Class 11 Chapter 15 \u0905\u0928\u0941\u0915\u094d\u0930\u092e \u0914\u0930 \u0936\u094d\u0930\u0947\u0923\u0940","og_description":"\u0936\u094d\u0930\u0947\u0923\u0940 15+12+9+... \u0915\u0947 \u0915\u093f\u0924\u0928\u0947 \u092a\u0926\u094b \u0915\u094b \u0932\u0947\u0928\u0947 \u0938\u0947 \u0909\u0928\u0915\u093e \u092f\u094b\u0917\u092b\u0932 15 \u092e\u093f\u0932\u0947\u0917\u093e? \u0926\u094b \u0909\u0924\u094d\u0924\u0930\u094b \u0915\u0940 \u0935\u094d\u092f\u093e\u0916\u094d\u092f\u093e \u0915\u0940\u091c\u093f\u090f\u0964 \u0936\u094d\u0930\u0947\u0923\u0940: 15+12+9+...","og_url":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-15-3-mathematics-solution-class-11-chapter-15-anukram-aur-shreni\/","og_site_name":"IndCareer Schools","article_publisher":"https:\/\/www.facebook.com\/indcareer","article_published_time":"2023-09-09T07:17:17+00:00","article_modified_time":"2023-09-09T07:17:22+00:00","og_image":[{"width":1600,"height":901,"url":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-16-1-scaled.jpg","type":"image\/jpeg"}],"author":"Pooja","twitter_card":"summary_large_image","twitter_creator":"@indcareer","twitter_site":"@indcareer","twitter_misc":{"Written by":"Pooja","Est. reading time":"6 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-15-3-mathematics-solution-class-11-chapter-15-anukram-aur-shreni\/#article","isPartOf":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-15-3-mathematics-solution-class-11-chapter-15-anukram-aur-shreni\/"},"author":{"name":"Pooja","@id":"https:\/\/www.indcareer.com\/schools\/#\/schema\/person\/d6945cf059726f162259ba738092301e"},"headline":"KC Sinha: Exercise 15.3 – Mathematics Solution Class 11 Chapter 15 \u0905\u0928\u0941\u0915\u094d\u0930\u092e \u0914\u0930 \u0936\u094d\u0930\u0947\u0923\u0940","datePublished":"2023-09-09T07:17:17+00:00","dateModified":"2023-09-09T07:17:22+00:00","mainEntityOfPage":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-15-3-mathematics-solution-class-11-chapter-15-anukram-aur-shreni\/"},"wordCount":1777,"publisher":{"@id":"https:\/\/www.indcareer.com\/schools\/#organization"},"image":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-15-3-mathematics-solution-class-11-chapter-15-anukram-aur-shreni\/#primaryimage"},"thumbnailUrl":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-16-1-scaled.jpg","articleSection":["class 11"],"inLanguage":"en-US"},{"@type":"WebPage","@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-15-3-mathematics-solution-class-11-chapter-15-anukram-aur-shreni\/","url":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-15-3-mathematics-solution-class-11-chapter-15-anukram-aur-shreni\/","name":"KC Sinha: Exercise 15.3 - Mathematics Solution Class 11 Chapter 15 \u0905\u0928\u0941\u0915\u094d\u0930\u092e \u0914\u0930 \u0936\u094d\u0930\u0947\u0923\u0940 - IndCareer Schools","isPartOf":{"@id":"https:\/\/www.indcareer.com\/schools\/#website"},"primaryImageOfPage":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-15-3-mathematics-solution-class-11-chapter-15-anukram-aur-shreni\/#primaryimage"},"image":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-15-3-mathematics-solution-class-11-chapter-15-anukram-aur-shreni\/#primaryimage"},"thumbnailUrl":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-16-1-scaled.jpg","datePublished":"2023-09-09T07:17:17+00:00","dateModified":"2023-09-09T07:17:22+00:00","description":"\u0936\u094d\u0930\u0947\u0923\u0940 15+12+9+... \u0915\u0947 \u0915\u093f\u0924\u0928\u0947 \u092a\u0926\u094b \u0915\u094b \u0932\u0947\u0928\u0947 \u0938\u0947 \u0909\u0928\u0915\u093e \u092f\u094b\u0917\u092b\u0932 15 \u092e\u093f\u0932\u0947\u0917\u093e? \u0926\u094b \u0909\u0924\u094d\u0924\u0930\u094b \u0915\u0940 \u0935\u094d\u092f\u093e\u0916\u094d\u092f\u093e \u0915\u0940\u091c\u093f\u090f\u0964 \u0936\u094d\u0930\u0947\u0923\u0940: 15+12+9+...","breadcrumb":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-15-3-mathematics-solution-class-11-chapter-15-anukram-aur-shreni\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-15-3-mathematics-solution-class-11-chapter-15-anukram-aur-shreni\/"]}]},{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-15-3-mathematics-solution-class-11-chapter-15-anukram-aur-shreni\/#primaryimage","url":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-16-1-scaled.jpg","contentUrl":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-16-1-scaled.jpg","width":1600,"height":901,"caption":"KC Sinha: Exercise 15.10 - Mathematics Solution Class 11 Chapter 15 \u0905\u0928\u0941\u0915\u094d\u0930\u092e \u0914\u0930 \u0936\u094d\u0930\u0947\u0923\u0940"},{"@type":"BreadcrumbList","@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-15-3-mathematics-solution-class-11-chapter-15-anukram-aur-shreni\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/www.indcareer.com\/schools\/"},{"@type":"ListItem","position":2,"name":"class 11","item":"https:\/\/www.indcareer.com\/schools\/class-11\/"},{"@type":"ListItem","position":3,"name":"KC Sinha: Exercise 15.3 – Mathematics Solution Class 11 Chapter 15 \u0905\u0928\u0941\u0915\u094d\u0930\u092e \u0914\u0930 \u0936\u094d\u0930\u0947\u0923\u0940"}]},{"@type":"WebSite","@id":"https:\/\/www.indcareer.com\/schools\/#website","url":"https:\/\/www.indcareer.com\/schools\/","name":"IndCareer Schools","description":"School Admissions & Notices","publisher":{"@id":"https:\/\/www.indcareer.com\/schools\/#organization"},"potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/www.indcareer.com\/schools\/?s={search_term_string}"},"query-input":{"@type":"PropertyValueSpecification","valueRequired":true,"valueName":"search_term_string"}}],"inLanguage":"en-US"},{"@type":"Organization","@id":"https:\/\/www.indcareer.com\/schools\/#organization","name":"IndCareer","url":"https:\/\/www.indcareer.com\/schools\/","logo":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/www.indcareer.com\/schools\/#\/schema\/logo\/image\/","url":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/06\/indcareer-logo2.png","contentUrl":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/06\/indcareer-logo2.png","width":512,"height":250,"caption":"IndCareer"},"image":{"@id":"https:\/\/www.indcareer.com\/schools\/#\/schema\/logo\/image\/"},"sameAs":["https:\/\/www.facebook.com\/indcareer","https:\/\/x.com\/indcareer","https:\/\/www.youtube.com\/channel\/UC1liU3RZoBRuu8YcAuZMsOQ"],"email":"info@ebharat.in","legalName":"IndCareer","numberOfEmployees":{"@type":"QuantitativeValue","minValue":"1","maxValue":"10"}},{"@type":"Person","@id":"https:\/\/www.indcareer.com\/schools\/#\/schema\/person\/d6945cf059726f162259ba738092301e","name":"Pooja","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/www.indcareer.com\/schools\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/350f7cfdfb6a23bcab67b56b5e77549db2a13b5d23e63175ac5bd07b5d44b720?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/350f7cfdfb6a23bcab67b56b5e77549db2a13b5d23e63175ac5bd07b5d44b720?s=96&d=mm&r=g","caption":"Pooja"}}]}},"_links":{"self":[{"href":"https:\/\/www.indcareer.com\/schools\/wp-json\/wp\/v2\/posts\/626028","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.indcareer.com\/schools\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.indcareer.com\/schools\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.indcareer.com\/schools\/wp-json\/wp\/v2\/users\/302"}],"replies":[{"embeddable":true,"href":"https:\/\/www.indcareer.com\/schools\/wp-json\/wp\/v2\/comments?post=626028"}],"version-history":[{"count":0,"href":"https:\/\/www.indcareer.com\/schools\/wp-json\/wp\/v2\/posts\/626028\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.indcareer.com\/schools\/wp-json\/wp\/v2\/media\/626030"}],"wp:attachment":[{"href":"https:\/\/www.indcareer.com\/schools\/wp-json\/wp\/v2\/media?parent=626028"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.indcareer.com\/schools\/wp-json\/wp\/v2\/categories?post=626028"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.indcareer.com\/schools\/wp-json\/wp\/v2\/tags?post=626028"},{"taxonomy":"boards","embeddable":true,"href":"https:\/\/www.indcareer.com\/schools\/wp-json\/wp\/v2\/boards?post=626028"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}