Sequence-And-Series Question 652
Question: Let S be the sum, P be the product and R be the sum of the reciprocals of 3 terms of a G.P. Then $ P^{2}R^{3}:S^{3} $ is equal to
Options:
A) 1 : 1
B) Common ratio: 1
C) $ {{( \text{first term} )}^{2}}\text{: }{{( \text{common ratio} )}^{2}} $
D) $ {{( commonratio )}^{n}}:1 $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] If the three terms of the GP. be $ \frac{a}{r}, $ a and or then $ S=\frac{a}{r}+a+ar=\frac{a}{r}(1+r+r^{2}) $ $ P=a^{3} $ and $ R=\frac{r}{a}+\frac{1}{a}+\frac{1}{ar}=\frac{1}{ar}(r^{2}+r+1) $ Now, $ \frac{P^{2}R^{3}}{S^{3}}=\frac{a^{6}\frac{1}{a^{3}r^{3}}{{(r^{2}+r+1)}^{3}}}{\frac{a^{3}}{r^{3}}{{(r^{2}+r+1)}^{3}}} $ So, the required ratio is $ 1:1 $ .
BETA
