Sequence And Series Question 592
Question: If the $ 5^{th} $ term of a G.P. is $ \frac{1}{3} $ and $ 9^{th} $ term is $ \frac{16}{243} $ , then the $ 4^{th} $ term will be
[MP PET 1982]
Options:
A) $ \frac{3}{4} $
B) $ \frac{1}{2} $
C) $ \frac{1}{3} $
D) $ \frac{2}{5} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ T_5=ar^{4}=\frac{1}{3} $ ?..(i) and $ T_9=ar^{8}=\frac{16}{243} $ ?..(ii) Solving (i) and (ii), we get $ r=\frac{2}{3} $ and $ a=\frac{27}{16} $ Now $ 4^{th} $ term $ =ar^{3}=\frac{3^{3}}{2^{4}}.\frac{2^{3}}{3^{3}}=\frac{1}{2} $ .
BETA
