8. Sequences and Series
hard

यदि  किसी गुणोत्तर श्रेणी के $n$ पदों का योग $S$ एवं गुणनफल $P$ है तथा उनके व्युत्क्रमों का योग $R$ है, तो ${P^2}$ का मान है  

A

$\frac{R}{S}$

B

$\frac{S}{R}$

C

${\left( {\frac{R}{S}} \right)^n}$

D

${\left( {\frac{S}{R}} \right)^n}$

(IIT-1966)

Solution

(d) दिया है,$S = \frac{{a({r^n} – 1)}}{{r – 1}} = \frac{{a\,(1 – {r^n})}}{{1 – r}}$……(i)

$P = a(ar)(a{r^2})……….(a{r^{n – 1}}) = {a^n}{r^{1 + 2 + ……… + (n – 1)}}$

$ = {a^n}{r^{(n – 1)n/2}}$, अर्थात् ${P^2} = {a^{2n}}{r^{n(n – 1)}}$ ……(ii)

एवं $R = \frac{1}{a} + \frac{1}{{ar}} + \frac{1}{{a{r^2}}} + ……….n$ पदों तक 

$ = \frac{1}{a}\left( {1 + \frac{1}{r} + \frac{1}{{{r^2}}} + ………\;n} \right)$ पदों तक 

$ = \frac{{\frac{1}{a}\left[ {{{\left( {\frac{1}{r}} \right)}^n} – 1} \right]}}{{\left( {\frac{1}{r} – 1} \right)}}$, $\left( {\because \,\frac{1}{r} > 1} \right.$ यदि $\left. \begin{array}{l}r < 1\\\end{array} \right)$

 $ = \frac{{(1 – {r^n})}}{{a{r^{n – 1}}(1 – r)}}$……. (iii)

अत:  $\frac{S}{R} = \frac{{a(1 – {r^n})}}{{1 – r}} \times \frac{{a{r^{n – 1}}(1 – r)}}{{(1 – {r^n})}} = {a^2}{r^{n – 1}}$

या ${\left( {\frac{S}{R}} \right)^n} = {({a^2}{r^{n – 1}})^n} = {a^{2n}}{r^{n(n – 1)}} = {P^2}$.

Standard 11
Mathematics

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