{"id":626037,"date":"2023-09-09T07:30:29","date_gmt":"2023-09-09T07:30:29","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=626037"},"modified":"2023-09-09T07:30:34","modified_gmt":"2023-09-09T07:30:34","slug":"kc-sinha-exercise-15-9-mathematics-solution-class-11-chapter-15-anukram-aur-shreni","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-15-9-mathematics-solution-class-11-chapter-15-anukram-aur-shreni\/","title":{"rendered":"KC Sinha: Exercise 15.9 – Mathematics Solution Class 11 Chapter 15 \u0905\u0928\u0941\u0915\u094d\u0930\u092e \u0914\u0930 \u0936\u094d\u0930\u0947\u0923\u0940"},"content":{"rendered":"\n

<\/p>\n\n\n\n

Question 1<\/h4>\n\n\n\n

2 \u0914\u0930 162 \u0915\u0947 \u092c\u0940\u091a 7 G.M’s \u0930\u093e\u0916\u093f\u090f\u0964
Sol :
\u092e\u093e\u0928\u093e 2 \u0914\u0930 162 \u0915\u0947 \u092c\u0940\u091a 7 G.M’s G1<\/sub>,G2<\/sub>,G3<\/sub>… G \u0939\u0948\u0964
2,G1<\/sub>,G2<\/sub>,G3<\/sub>,…G7<\/sub>,162 G.P \u092e\u0947 \u0939\u0948\u0964<\/p>\n\n\n\n

a=2 ,b=162, n=7<\/p>\n\n\n\n

$r=\\left(\\frac{b}{a}\\right)^{\\frac{1}{n+1}}$
$=\\left(\\frac{162}{2}\\right)^{\\frac{1}{7+1}}$
$=3^{4\\times \\frac{1}{8}}=3^{\\frac{1}{2}}$
=\u221a3<\/p>\n\n\n\n

G1<\/sub>=ar=2\u221a3<\/p>\n\n\n\n

G2<\/sub>=ar2<\/sup>=2(\u221a3)2<\/sup>=6<\/p>\n\n\n\n

G3<\/sub>=ar3<\/sup>=2(\u221a3)3<\/sup>
=2\u00d73\u221a3=6\u221a3<\/p>\n\n\n\n

G4<\/sub>=ar4<\/sup>=2(\u221a3)4<\/sup>
=2\u00d79=18<\/p>\n\n\n\n

G5<\/sub>=ar5<\/sup>=2(\u221a3)5<\/sup>
=2\u00d79\u221a3=18\u221a3<\/p>\n\n\n\n

G6<\/sub>=ar6<\/sup>=2(\u221a3)6<\/sup>
=2\u00d727=54<\/p>\n\n\n\n

G7<\/sub>=ar7<\/sup>=2(\u221a3)7<\/sup>
=2\u00d727\u221a3=54\u221a3<\/p>\n\n\n\n

2 \u0914\u0930 162 \u0915\u0947 \u092c\u0940\u091a \u0915\u0947 7 G.M’s<\/p>\n\n\n\n

2\u221a3,6,6\u221a3,18,18\u221a3,54,54\u221a3<\/p>\n\n\n\n

Question 3<\/h4>\n\n\n\n

\u092f\u0926\u093f \u0926\u094b \u0927\u0928\u093e\u0924\u094d\u092e\u0915 \u0938\u0902\u0916\u094d\u092f\u093e\u0913 a \u0924\u0925\u093e b \u0915\u0947 \u092c\u0940\u091a \u0938\u092e\u093e\u0902\u0924\u0930 \u092e\u093e\u0927\u094d\u092f \u0924\u0925\u093e \u0917\u0941\u0923\u094b\u0924\u094d\u0924\u0930 \u092e\u093e\u0927\u094d\u092f \u0915\u094d\u0930\u092e\u0936\u0903 10 \u0924\u0925\u093e 8 \u0939\u0948 , \u0924\u094b \u0938\u0902\u0916\u094d\u092f\u093e\u090f\u0901 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964
Sol :
$\\frac{a+b}{2}=10$ , \u221aab=8<\/p>\n\n\n\n

a+b=20..(i)
\u0926\u094b\u0928\u094b \u0924\u0930\u0923 \u0935\u0930\u094d\u0917 \u0915\u0930\u0928\u0947 \u092a\u0930,
(\u221aab)2<\/sup>=82<\/sup>
ab=64
$b=\\frac{64}{a}$<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u092e\u0947 b \u0915\u093e \u092e\u093e\u0928 \u0930\u0916\u0928\u0947 \u092a\u0930,
a+b=20
$a+\\frac{64}{a}=20$
$\\frac{a^2+64}{a}=20$
a2<\/sup>+64=200
a2<\/sup>-20a+64=0
a(a-16)-4(a-16)=0
(a-16)(a-4)=0<\/p>\n\n\n\n

$\\begin{array}{l|l}a-16=0&a-4=0\\\\a=16&a=4\\end{array}$<\/p>\n\n\n\n

CASE-I
a=16<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u0938\u0947
a+b=20
16+b=20
b=4<\/p>\n\n\n\n

CASE-II
a=4
\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u0938\u0947
a+b=20
4+b=20
b=16<\/p>\n\n\n\n

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

\u0926\u094b-\u0926\u0940 \u0917\u0908 \u0930\u093e\u0936\u093f\u092f\u094b a \u0924\u0925\u093e b \u0915\u0947 \u092c\u0940\u091a \u0935\u093f\u0937\u092e \u0938\u0902\u0916\u094d\u092f\u093e\u0913 \u092e\u0947 G.M’s \u0930\u0916\u0947 \u0917\u092f\u094b \u0939\u094b \u0924\u094b \u0926\u093f\u0916\u0932\u093e\u0907\u090f \u0915\u093f \u092e\u0927\u094d\u092f G.M=\u221aab
Sol :
\u092e\u093e\u0928\u093e \u0926\u094b-\u0926\u0940 \u0917\u0908 \u0930\u093e\u0936\u093f\u092f\u094b a \u0924\u0925\u093e b \u0915\u0947 \u092c\u0940\u091a<\/p>\n\n\n\n

(2n+1) G.M’s \u0930\u0916\u0947 \u0917\u092f\u0947 \u0939\u0948\u0964<\/p>\n\n\n\n

\u092e\u093e\u0928\u093e G.P \u0915\u093e \u0938\u093e\u0930\u094d\u0935 \u0905\u0928\u0941\u092a\u093e\u0924 r \u0939\u0948\u0964<\/p>\n\n\n\n

$r=\\left(\\frac{b}{a}\\right)^{\\frac{1}{2n+1+1}}$<\/p>\n\n\n\n

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

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

G.P \u0915\u093e \u092e\u0927\u094d\u092f \u092a\u0930 \u092f\u093e \u0930\u0916\u0947 \u0917\u090f (2n+1) \u092a\u0926\u094b \u092e\u0947 \u092e\u0927\u094d\u092f \u0915\u093e
G.M=$\\left(\\frac{2n}{2}+1\\right)$ \u0935\u093e\u0901 \u092a\u0926
=(n+1)\u0935\u093e\u0901 \u092a\u0926<\/p>\n\n\n\n

Gn+1<\/sub>=arn+1<\/sup><\/p>\n\n\n\n

$=a\\left(\\frac{b}{a}\\right)^{(n+1)\\times \\frac{1}{2(n+1)}}$<\/p>\n\n\n\n

$=a.\\dfrac{b^{\\frac{1}{2}}}{a^{\\frac{1}{2}}}$<\/p>\n\n\n\n

$=a^{\\frac{1}{2}}b^{\\frac{1}{2}}$<\/p>\n\n\n\n

$=(ab)^{\\frac{1}{2}}$
=\u221aab<\/p>\n\n\n\n

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

\u092f\u0926\u093f \u0926\u094b \u0938\u0902\u0916\u094d\u092f\u093e\u0913 \u0915\u0947 \u092c\u0940\u091a A.M \u0924\u0925\u093e G.M \u0915\u094d\u0930\u092e\u0936\u0903 A \u0914\u0930 G \u0939\u094b \u0924\u094b \u0926\u093f\u0916\u0932\u093e\u0907\u090f \u0915\u093f \u0935\u0947 \u0938\u0902\u0916\u094d\u092f\u093e\u090f\u0901 A+$\\sqrt{A^2-G^2}$ \u0914\u0930 $A-\\sqrt{A^2-G^2}$ \u0939\u0948\u0964
Sol :
\u092e\u093e\u0928\u093e \u0935\u0939 \u0926\u094b \u0938\u0902\u0916\u094d\u092f\u093e\u090f\u0901 a \u0924\u0925\u093e b \u0939\u0948\u0964<\/p>\n\n\n\n

$\\frac{a+b}{2}=A$ , \u221aab=G<\/p>\n\n\n\n

\u0926\u094b\u0928\u094b \u0915\u094b \u0935\u0930\u094d\u0917 \u0915\u0930\u0915\u0947 \u0918\u091f\u093e\u0928\u0947 \u092a\u0930,<\/p>\n\n\n\n

$\\left(\\frac{a+b}{2}\\right)^2$-(\u221aab)2<\/sup>=A2<\/sup>-G2<\/sup><\/p>\n\n\n\n

$\\frac{a^2+b^2+2ab}{4}$-ab=A2<\/sup>-G2<\/sup><\/p>\n\n\n\n

$\\frac{a^2+b^2-4ab}{4}$=A2<\/sup>-G2<\/sup><\/p>\n\n\n\n

$\\frac{a^2+b^2-2ab}{4}$=A2<\/sup>-G2<\/sup><\/p>\n\n\n\n

$\\left(\\frac{a-b}{2}\\right)^2$=A2<\/sup>-G2<\/sup><\/p>\n\n\n\n

$\\frac{a-b}{2}=\\pm \\sqrt{A^2-G^2}$<\/p>\n\n\n\n

CASE-I
$\\frac{a+b}{2}=A$ , $\\frac{a-b}{2}=\\sqrt{A^2-G^2}$<\/p>\n\n\n\n

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

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

$\\frac{2a}{2}=A+\\sqrt{A^2-G^2}$<\/p>\n\n\n\n

a=A+\u221aA2<\/sup>-G2<\/sup><\/p>\n\n\n\n

\u0918\u091f\u093e\u0928\u0947 \u092a\u0930,
$\\frac{a+b}{2}-\\frac{a-b}{2}=A-\\sqrt{A^2-G^2}$<\/p>\n\n\n\n

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

$\\frac{2b}{2}=A-\\sqrt{A^2-G^2}$<\/p>\n\n\n\n

$b=A-\\sqrt{A^2-G^2}$<\/p>\n\n\n\n

CASE-II
$\\frac{a+b}{2}=A$ , $\\frac{a-b}{2}=-\\sqrt{A^2-G^2}$<\/p>\n\n\n\n

\u091c\u094b\u0921\u093c\u0928\u0947 \u092a\u0930,
$\\frac{a+b}{2}+\\frac{a-b}{2}=A-\\sqrt{A^2-G^2}$<\/p>\n\n\n\n

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

$\\frac{2a}{a}=A-\\sqrt{A^2-G^2}$<\/p>\n\n\n\n

$a=A-\\sqrt{A^2-G^2}$<\/p>\n\n\n\n

\u0918\u091f\u093e\u0928\u0947 \u092a\u0930,
$\\frac{a+b}{2}-\\frac{a-b}{2}=A-\\left(-\\sqrt{A^2-G^2}\\right)$<\/p>\n\n\n\n

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

$\\frac{2b}{2}=A+\\sqrt{A^2-G^2}$<\/p>\n\n\n\n

$a=A+\\sqrt{A^2-G^2}$<\/p>\n\n\n\n

\u0905\u0924\u0903 \u0905\u092d\u093f\u0937\u094d\u091f \u0930\u093e\u0936\u093f\u092f\u093e\u0901,
$A+\\sqrt{A^2-G^2},A-\\sqrt{A^2-G^2}$<\/p>\n\n\n\n

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

\u092f\u0926\u093f \u0926\u094b \u0938\u0902\u0916\u094d\u092f\u093e\u0913 a \u0914\u0930 b \u0915\u0947 \u092c\u0940\u091a \u0915\u0947 A.M \u0924\u0925\u093e G.M \u0915\u093e \u0905\u0928\u0941\u0935\u093e\u0926 m:n \u0939\u094b \u0924\u094b \u0938\u093e\u092c\u093f\u0924 \u0915\u0940\u091c\u093f\u090f \u0915\u093f
a:b=m+$\\sqrt{m^2-n^2}:m-\\sqrt{m^2-n^2}$
Sol :
a \u0914\u0930 b \u0915\u0947 \u092c\u0940\u091a \u0915\u0947 A.M \u0924\u0925\u093e G.M \u0915\u093e \u0905\u0928\u0941\u092a\u093e\u0924=m:n<\/p>\n\n\n\n

$\\dfrac{\\frac{a+b}{2}}{\\sqrt{ab}}=\\frac{m}{n}$<\/p>\n\n\n\n

$\\frac{a+b}{2\\sqrt{ab}}=\\frac{m}{n}$<\/p>\n\n\n\n

By componendo and dividendo<\/p>\n\n\n\n

$\\frac{a+b+2\\sqrt{ab}}{a+b-2\\sqrt{ab}}=\\frac{m+n}{m-n}$<\/p>\n\n\n\n

$\\frac{(\\sqrt{a})^2+(\\sqrt{b})^2+2\\sqrt{a}\\sqrt{b}}{(\\sqrt{a})^2+(\\sqrt{b})^2-2\\sqrt{a}\\sqrt{b}}=\\frac{m+n}{m-n}$<\/p>\n\n\n\n

$\\frac{(\\sqrt{a}+\\sqrt{b})^2}{(\\sqrt{a}-\\sqrt{b})^2}=\\frac{m+n}{m-n}$<\/p>\n\n\n\n

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

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

Again , by componendo and dividendo<\/p>\n\n\n\n

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

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

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

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

\u0926\u094b\u0928\u094b \u0924\u0930\u092b \u0935\u0930\u094d\u0917 \u0915\u0930\u0928\u0947 \u092a\u0930,<\/p>\n\n\n\n

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

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

$\\frac{a}{b}=\\frac{(\\sqrt{m+n})^2+(\\sqrt{m-n})^2+2\\sqrt{m+n}.\\sqrt{m-n}}{(\\sqrt{m+n})^2+(\\sqrt{m-n})^2-2\\sqrt{m+n}\\sqrt{m-n}}$<\/p>\n\n\n\n

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

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

$\\frac{a}{b}=\\frac{2(m+\\sqrt{m^2-n^2})}{2(m-\\sqrt{m^2-n^2)}}$<\/p>\n\n\n\n

a:b=(m+\u221am2<\/sup>+n2<\/sup>):(m-\u221am2<\/sup>-n2<\/sup>)<\/p>\n\n\n\n

Question 9<\/h4>\n\n\n\n

\u092f\u0926\u093f \u0926\u094b \u0926\u0940 \u0917\u0908 \u0930\u093e\u0936\u093f\u0913 a \u0924\u0925\u093e b \u0915\u0947 \u092c\u0940\u091a \u090f\u0915 G.M.G \u0914\u0930 \u0926\u094b A.M’s p \u0924\u0925\u093e q \u0930\u0916\u0947 \u0928\u092f\u0947 \u0939\u094b \u0924\u094b \u0938\u093e\u092c\u093f\u0924 \u0915\u0940\u091c\u093f\u090f \u0915\u093f
G2<\/sup>=(2p-q)(2p-p)
Sol :
a,G,b, G.P \u092e\u0947 \u0939\u0948\u0964<\/p>\n\n\n\n

$\\frac{G}{a}=\\frac{b}{G}$<\/p>\n\n\n\n

G2<\/sup>=ab..(i)<\/p>\n\n\n\n

a,p,q,b A.P \u092e\u0947 \u0939\u0948\u0964<\/p>\n\n\n\n

\u092e\u093e\u0928\u093e \u0907\u0938 A.P \u0915\u093e \u0938\u093e\u0930\u094d\u0935\u0905\u0902\u0924\u0930 d \u0939\u0948 \u0964<\/p>\n\n\n\n

$d=\\frac{b-a}{2+1}=\\frac{b-a}{3}$<\/p>\n\n\n\n

p=a+d
$=a+\\frac{b-a}{3}$
$=\\frac{3a+b-a}{3}$
$=\\frac{2a+b}{3}$<\/p>\n\n\n\n

$P=\\frac{2a+b}{3}$
2a+b=3p..(i)<\/p>\n\n\n\n

q=a+2d
$=a+2\\left(\\frac{b-a}{3}\\right)$
$=\\frac{3a+2b-2a}{3}$
$q=\\frac{a+2b}{3}$
a+2b=3q..(ii)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u092e\u0947 2 \u0938\u0947 \u0924\u0925\u093e \u0938\u092e\u0940\u0915\u0930\u0923 (ii) \u092e\u0947 1 \u0938\u0947 \u0917\u0941\u0923\u093e \u0915\u0930\u0928\u0947 \u092a\u0930,<\/p>\n\n\n\n

$\\begin{aligned}4a+2b=&6p\\\\a+2b=&3q\\\\-\\phantom{a}-\\phantom{2b}&-\\phantom{3q}\\\\ 3a=6p-3q\\end{aligned}$<\/p>\n\n\n\n

3a=3(2p-q)
a=2p-q<\/p>\n\n\n\n

a \u0915\u093e \u092e\u093e\u0928 \u0938\u092e\u0940\u0915\u0930\u0923 (ii) \u092e\u0947 \u0930\u0916\u0928\u0947 \u092a\u0930,
a+2b=3q
2p-q+2b=3q
2b=3q-2p+q
2b=4q-2p
2b=2(2q-p)
b=(2q-p)<\/p>\n\n\n\n

\u2235G2<\/sup>=ab=(2p-q)(2q-p)<\/p>\n\n\n\n

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

\u092f\u0926\u093f \u0926\u094b \u0926\u0940 \u0917\u0908 \u0938\u0902\u0916\u094d\u092f\u093e\u0913 \u0915\u0947 \u092c\u0940\u091a \u090f\u0915 A.M, A \u0914\u0930 \u0926\u094b G.M’s p \u0924\u0925\u093e q \u0930\u0916\u0947 \u0917\u092f\u0947 \u0939\u094b \u0924\u094b \u0926\u093f\u0916\u0932\u093e\u0907\u090f \u0915\u093f
$\\frac{p^2}{q}+\\frac{q^2}{p}=2A$
Sol :
\u092e\u093e\u0928\u093e a \u0914\u0930 b \u0926\u094b \u0926\u0940 \u0917\u0908 \u0938\u0902\u0916\u092f\u093e\u090f\u0901 \u0939\u0948\u0964
a,A,b A.P \u092e\u0947 \u0939\u0948 \u0964<\/p>\n\n\n\n

$A=\\frac{a+b}{2}$<\/p>\n\n\n\n

2A=a+b..(i)<\/p>\n\n\n\n

a,p,q,b G.P \u092e\u0947 \u0939\u0948 \u0964<\/p>\n\n\n\n

\u092e\u093e\u0928\u093e G.P \u0915\u093e \u0938\u093e\u0930\u094d\u0935 \u0905\u0928\u0941\u092a\u093e\u0924 r \u0939\u0948\u0964<\/p>\n\n\n\n

$r=\\left(\\frac{b}{a}\\right)^{\\frac{1}{2+1}}$<\/p>\n\n\n\n

$=\\left(\\frac{b}{a}\\right)^{\\frac{1}{3}}$<\/p>\n\n\n\n

p=ar
$=a\\left(\\frac{b}{a}\\right)^{\\frac{1}{3}}$<\/p>\n\n\n\n

$=a.\\dfrac{b^{\\frac{1}{3}}}{a^{\\frac{1}{3}}}$<\/p>\n\n\n\n

$=a^{\\frac{2}{3}}.b^{\\frac{1}{3}}$<\/p>\n\n\n\n

q=ar2<\/sup>
$=a\\left(\\frac{b}{a}\\right)^{\\frac{2}{3}}$<\/p>\n\n\n\n

$=a\\dfrac{b^{\\frac{2}{3}}}{a^{\\frac{2}{3}}}$<\/p>\n\n\n\n

$=a^{\\frac{1}{3}}.b^{\\frac{2}{3}}$<\/p>\n\n\n\n

L.H.S
$\\frac{p^2}{q}+\\frac{q^2}{p}$<\/p>\n\n\n\n

$=\\dfrac{\\left(a^{\\frac{2}{3}}.b^{\\frac{1}{3}}\\right)^2}{a^{\\frac{1}{3}}.b^{\\frac{2}{3}}}+\\dfrac{\\left(a^{\\frac{1}{3}}.b^{\\frac{2}{3}}\\right)^2}{a^{\\frac{2}{3}}.b^{\\frac{1}{3}}}$<\/p>\n\n\n\n

$=\\dfrac{a^{\\frac{4}{3}.b^{\\frac{2}{3}}}}{a^{\\frac{1}{3}}.b^{\\frac{2}{3}}}+\\dfrac{a^{\\frac{2}{3}.b^{\\frac{4}{3}}}}{a^{\\frac{2}{3}}.b^{\\frac{1}{3}}}$<\/p>\n\n\n\n

$=a^{\\frac{3}{3}}+b^{\\frac{3}{3}}$<\/p>\n\n\n\n

=a+b (\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u0938\u0947 )<\/p>\n\n\n\n

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

\u092f\u0926\u093f $0<\\theta <\\frac{\\pi}{2}$ , \u0924\u094b tan\u03b8+cot\u03b8 \u0915\u093e \u0928\u094d\u092f\u0942\u0928\u0924\u092e \u092e\u093e\u0928 \u0928\u093f\u0915\u093e\u0932\u0947\u0964
Sol :
0<2\u03b8<\u03c0<\/p>\n\n\n\n

tan\u03b8+cot\u03b8$=\\frac{\\sin \\theta}{\\cos \\theta}+\\frac{\\cos \\theta}{\\sin \\theta}$<\/p>\n\n\n\n

$=\\frac{\\sin^2 \\theta+\\cos^2 \\theta}{\\cos \\theta \\sin \\theta}$<\/p>\n\n\n\n

$=\\frac{1}{\\sin \\theta \\cos \\theta}$<\/p>\n\n\n\n

$=\\frac{2}{2\\sin \\theta \\cos \\theta}=\\frac{2}{\\sin 2\\theta}$<\/p>\n\n\n\n

=2cosec2\u03b8
=2cosec2\u03b8\u22652<\/p>\n\n\n

[cosec2\u03b8>0
\n0<2\u03b8<\u03c0]<\/p>\n\n\n\n

\u0928\u094d\u092f\u0942\u0928\u0924\u092e \u092e\u093e\u0928=2<\/p>\n\n\n\n

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

\u092f\u0926\u093f \u0915\u093f\u0938\u0940 A.P \u0924\u0925\u093e G.P \u0926\u094b\u0928\u094b \u0915\u093e p \u0935\u093e\u0901, q \u0935\u093e\u0901 \u0924\u0925\u093e r \u0935\u093e\u0901 \u092a\u0926 a,b,c \u0939\u094b \u0924\u094b \u0938\u093e\u092c\u093f\u0924 \u0915\u0940\u091c\u093f\u090f \u0915\u093f
ab<\/sup>bc<\/sup>ca<\/sup>=ac<\/sup>ba<\/sup>cb<\/sup>
Sol :
\u092e\u093e\u0928\u093e A.P \u0915\u093e \u092a\u094d\u0930\u0925\u092e \u092a\u0926 A1<\/sub> \u0924\u0925\u093e \u0938\u093e\u0930\u094d\u0935 \u0905\u0928\u0941\u092a\u093e\u0924 D \u0939\u0948 \u0964<\/p>\n\n\n\n

A1<\/sub>+(p-1)D=a..(i)
A1<\/sub>+(q-1)D=b..(ii)
A1<\/sub>+(r-1)D=c..(iii)<\/p>\n\n\n\n

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

$\\begin{aligned}A_1+(p-1)D=&a\\\\A_1+(q-1)D=&b\\\\-\\phantom{A_1}-\\pahntom{(q-1)D}&-\\phantom{b}\\\\(p-1)D-(q-1)D=a-b\\end{aligned}$<\/p>\n\n\n\n

(p-1-q+1)D=a-b
(p-q)D=a-b<\/p>\n\n\n\n

\u0907\u0938\u0940 \u092a\u094d\u0930\u0915\u093e\u0930,
(q-r)D=b-c
(r-p)D=c-a<\/p>\n\n\n\n

\u092e\u093e\u0928\u093e G.P \u0915\u093e \u092a\u094d\u0930\u0925\u092e \u092a\u0926 A \u0924\u0925\u093e \u0938\u093e\u0930\u094d\u0935 \u0905\u0928\u0941\u092a\u093e\u0924 R \u0939\u0948\u0964<\/p>\n\n\n\n

ARp-1<\/sup>=a..(iv)
ARq-1<\/sup>=b…(v)
ARr-1<\/sup>=c…(vi)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (iv) \u0915\u093e \u0918\u093e\u0924 (b-c), \u0938\u092e\u0940\u0915\u0930\u0923 (v) \u0915\u093e \u0918\u093e\u0924 (c-a) \u0924\u0925\u093e \u0938\u092e\u0940\u0915\u0930\u0923 (vi) \u0915\u093e \u0918\u093e\u0924 (a-b) \u0932\u0947\u0915\u0930 \u0924\u0940\u0928\u094b \u0938\u092e\u0940\u0915\u0930\u0923 \u092e\u0947 \u0917\u0941\u0923\u093e \u0915\u0930\u0928\u0947 \u092a\u0930,<\/p>\n\n\n\n

(ARp-1<\/sup>)(b-c),(ARq-1<\/sup>)(c-a),(ARr-1<\/sup>)(a-b)=ab-c<\/sup>,bc-a<\/sup>,ca-b<\/sup>
(ARp-1<\/sup>)(q-r)D,(ARq-1<\/sup>)(r-p)D,(ARr-1<\/sup>)(p-q)D=ab-c<\/sup>,bc-a<\/sup>,ca-b<\/sup><\/p>\n\n\n\n

[Aq-r<\/sup>.Rpq-pr-q+r<\/sup>.Ar-p<\/sup>.Rqr-pq-r+p<\/sup>.Ap-q<\/sup>.Rpr-qr-p+q<\/sup>]=ab-c<\/sup>.bc-a<\/sup>.ca-b<\/sup><\/p>\n\n\n\n

A0<\/sup>R0<\/sup>=ab-c<\/sup>.bc-a<\/sup>.ca-b<\/sup><\/p>\n\n\n\n

$1=\\frac{a^b}{a^c}\\times \\frac{b^c}{b^a} \\times \\frac{c^a}{c^b}$<\/p>\n\n\n\n

ac<\/sup>ba<\/sup>cb<\/sup>=ab<\/sup>bc<\/sup>ca<\/sup><\/p>\n\n\n\n

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

\u0927\u0928\u093e\u0924\u094d\u092e\u0915 \u092a\u0926\u094b \u0935\u093e\u0932\u0947 \u090f\u0915 A.P \u0924\u0925\u093e \u090f\u0915 G.P \u0915\u093e \u092a\u094d\u0930\u0925\u092e \u092a\u0926 \u0939\u0948 \u0914\u0930 \u0909\u0928\u0915\u0947 \u092a\u094d\u0930\u0925\u092e, \u0926\u094d\u0935\u093f\u0924\u0940\u092f \u0924\u0925\u093e \u0924\u0943\u0924\u0940\u092f \u092a\u0926\u094b \u0915\u093e \u092f\u094b\u0917 \u0915\u094d\u0930\u092e\u0936\u0903 $1,\\frac{1}{2}$ \u0914\u0930 2 \u0939\u0948\u0964 \u0909\u0928\u0915\u0947 \u091a\u0924\u0941\u0930\u094d\u0925 \u092a\u0926\u094b \u0915\u093e \u091c\u094b\u0921\u093c \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964
Sol :
\u092e\u093e\u0928\u093e A.P \u0924\u0925\u093e G.P \u0926\u094b\u0928\u094b \u0915\u0947 \u092a\u094d\u0930\u0925\u092e \u092a\u0926 a \u0939\u0948\u0964<\/p>\n\n\n\n

A.P \u0915\u093e \u0938\u093e\u0930\u094d\u0935\u0905\u0928\u0941\u092a\u093e\u0924 d \u0924\u0925\u093e G.P \u0915\u093e \u0938\u093e\u0930\u094d\u0935\u0905\u0928\u0941\u092a\u093e\u0924 r \u0939\u0948\u0964<\/p>\n\n\n\n

A.P: a,a+d,a+2d,a+3d…<\/p>\n\n\n\n

G.P: a,ar,ar2<\/sup>,ar3<\/sup>….<\/p>\n\n\n\n

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

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

$a=\\frac{1}{2}$ \u092a\u0930,<\/p>\n\n\n\n

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

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

a+2d+ar2<\/sup>=2<\/p>\n\n\n\n

$a=\\frac{1}{2}$ \u092a\u0930,<\/p>\n\n\n\n

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

$2d+\\frac{1}{2}r^2=2-\\frac{1}{2}$<\/p>\n\n\n\n

$2\\left(-\\frac{1}{2}r\\right)+\\frac{1}{2}r^2=\\frac{4-1}{2}$<\/p>\n\n\n\n

$-r+\\frac{1}{2}r^2=\\frac{3}{2}$<\/p>\n\n\n\n

$\\frac{-2r+r^2}{2}=\\frac{3}{2}$<\/p>\n\n\n\n

r2<\/sup>-2r-3=0
r2<\/sup>-3r+r-3=0
r(r-3)+1(r-3)=0
(r-3)(r+1)=0<\/p>\n\n\n\n

$\\begin{array}{l|l}r-3=0&r+1=0\\\\r=3&r=-1\\end{array}$<\/p>\n\n\n\n

r  \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

r=3 \u092a\u0930,
$d=-\\frac{1}{2}(3)=-\\frac{3}{2}$<\/p>\n\n\n\n

r=1 \u092a\u0930,
$d=-\\frac{1}{2}(-1)=-\\frac{1}{2}$<\/p>\n\n\n\n

A.P  \u0914\u0930 G.P \u0915\u0947 \u091a\u0924\u0941\u0930\u094d\u0925 \u092a\u0926\u094b \u0915\u093e \u092f\u094b\u0917\u0903
(a+3d)+ar3<\/sup>=?<\/p>\n\n\n\n

CASE-I
$a=\\frac{1}{2}$ , r=3, $d=-\\frac{3}{2}$<\/p>\n\n\n\n

(a+3d)+ar3<\/sup>=$\\frac{1}{2}+3\\left(-\\frac{3}{2}\\right)+\\frac{1}{2}(3)^3$<\/p>\n\n\n\n

$=\\frac{1}{2}-\\frac{9}{2}+\\frac{27}{2}$<\/p>\n\n\n\n

$=\\frac{1-9+27}{2}=\\frac{19}{2}$<\/p>\n\n\n\n

CASE-II
$a=\\frac{1}{2}$, r=-1, $d=\\frac{1}{2}$<\/p>\n\n\n\n

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

$=\\frac{1}{2}+\\frac{3}{2}-\\frac{1}{2}$<\/p>\n\n\n\n

$\\frac{3}{2}$<\/p>\n\n\n\n

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

Question 1 2 \u0914\u0930 162 \u0915\u0947 \u092c\u0940\u091a 7 G.M’s \u0930\u093e\u0916\u093f\u090f\u0964Sol :\u092e\u093e\u0928\u093e 2 \u0914\u0930 162 \u0915\u0947 \u092c\u0940\u091a 7 G.M’s G1,G2,G3… G \u0939\u0948\u09642,G1,G2,G3,…G7,162 G.P \u092e\u0947 \u0939\u0948\u0964 a=2 ,b=162, n=7 $r=\\left(\\frac{b}{a}\\right)^{\\frac{1}{n+1}}$$=\\left(\\frac{162}{2}\\right)^{\\frac{1}{7+1}}$$=3^{4\\times \\frac{1}{8}}=3^{\\frac{1}{2}}$=\u221a3 G1=ar=2\u221a3 G2=ar2=2(\u221a3)2=6 G3=ar3=2(\u221a3)3=2\u00d73\u221a3=6\u221a3 G4=ar4=2(\u221a3)4=2\u00d79=18 G5=ar5=2(\u221a3)5=2\u00d79\u221a3=18\u221a3 G6=ar6=2(\u221a3)6=2\u00d727=54 G7=ar7=2(\u221a3)7=2\u00d727\u221a3=54\u221a3 2 \u0914\u0930 162 \u0915\u0947 \u092c\u0940\u091a \u0915\u0947 7 G.M’s 2\u221a3,6,6\u221a3,18,18\u221a3,54,54\u221a3 Question 3 \u092f\u0926\u093f \u0926\u094b \u0927\u0928\u093e\u0924\u094d\u092e\u0915 \u0938\u0902\u0916\u094d\u092f\u093e\u0913 a \u0924\u0925\u093e b \u0915\u0947 […]<\/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-626037","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-class-11","entry"],"yoast_head":"\nKC Sinha: Exercise 15.9 - 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=\"Question 1 2 \u0914\u0930 162 \u0915\u0947 \u092c\u0940\u091a 7 G.M's \u0930\u093e\u0916\u093f\u090f\u0964Sol :\u092e\u093e\u0928\u093e 2 \u0914\u0930 162 \u0915\u0947 \u092c\u0940\u091a 7 G.M's G1,G2,G3... 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