Sequence And Series Question 142
Question: If the sum of the $ n $ terms of G.P. is $ S $ product is $ P $ and sum of their inverse is $ R $ , than $ P^{2} $ is equal to
[IIT 1966; Roorkee 1981]
Options:
A) $ \frac{R}{S} $
B) $ \frac{S}{R} $
C) $ {{( \frac{R}{S} )}^{n}} $
D) $ {{( \frac{S}{R} )}^{n}} $
Show Answer
Answer:
Correct Answer: D
Solution:
Given that sum $ S=\frac{a(r^{n}-1)}{r-1}=\frac{a,(1-r^{n})}{1-r} $ ……(i) $ P=a(ar)(ar^{2})……….(a{r^{n-1}})=a^{n}{r^{1+2+………+(n-1)}} $ $ =a^{n}{r^{(n-1)n/2}}\ i.e.,\ P^{2}=a^{2n}{r^{n(n-1)}} $ ……(ii) and $ R=\frac{1}{a}+\frac{1}{ar}+\frac{1}{ar^{2}}+………. $ upto $ \frac{1}{49} $ terms $ =\frac{1}{a}( 1+\frac{1}{r}+\frac{1}{r^{2}}+………upto\ n\ terms ) $ $ =\frac{\frac{1}{a}[ {{( \frac{1}{r} )}^{n}}-1 ]}{( \frac{1}{r}-1 )}( \because \ \frac{1}{r}>1 ) $ if $ r<1 $ $ =\frac{(1-r^{n})}{a{r^{n-1}}(1-r)} $ ……. (iii) Therefore , $ \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}} $ or $ {{( \frac{S}{R} )}^{n}}={{(a^{2}{r^{n-1}})}^{n}}=a^{2n}{r^{n(n-1)}}=P^{2} $ .
BETA
