Sequence And Series Question 520
Question: If $ \frac{3+5+7+……….to\ n\ terms}{5+8+11+………to\ 10\ terms}=7 $ , then the value of $ n $ is
[MNR 1983; Pb. CET 2000]
Options:
A) 35
B) 36
C) 37
D) 40
Show Answer
Answer:
Correct Answer: A
Solution:
We have $ \frac{3+5+7+……upto\ n\ terms}{5+8+11+……..upto\ 10\ terms}=7 $
$ \Rightarrow \frac{\frac{n}{2}[6+(n-1)2]}{\frac{10}{2}[10+(10-1)3]}=7\Rightarrow \frac{n(2n+4)}{10\times 37}=7 $
$ \Rightarrow $ $ n^{2}+2n-1295=0\Rightarrow (n+37)(n-35)=0 $ Hence $ n=35 $ .
BETA
