Sequence And Series Question 211
Question: If the ratio of the sum of $ n $ terms of two A.P.’s be $ (7n+1):(4n+27) $ , then the ratio of their $ 11^{th} $ terms will be
[AMU 1996]
Options:
A) $ 2:3 $
B) $ 3:4 $
C) $ 4:3 $
D) $ 5:6 $
Show Answer
Answer:
Correct Answer: C
Solution:
Let $ S_{n} $ and $ S{’{n}} $ be the sums of n terms of two A.P.’s and $ T{11} $ and $ T{’{11}} $ be the respective $ 11^{th} $ terms, then $ \frac{S{n}}{S{’{n}}}=\frac{\frac{n}{2}[2a+(n-1)d]}{\frac{n}{2}[2a’+(n-1)d’]}=\frac{7n+1}{4n+27} $
$ \Rightarrow $ $ \frac{a+\frac{(n-1)}{2}d}{a’+\frac{(n-1)}{2}d’}=\frac{7n+1}{4n+27} $ Now put $ n=21 $ , we get $ \frac{a+10d}{a’+10d’}=\frac{T{11}}{T{’_{11}}}=\frac{148}{111}=\frac{4}{3} $ . Note : If ratio of sum of $ n $ terms of two A.P.’s are given in terms of $ n $ and ratio of their $ p^{th} $ terms are to be found then put $ n=2p-1 $ . Here we put $ n=11\times 2-1=21 $ .
BETA
