Sequence And Series Question 515
Question: The number of terms of the A.P. 3,7,11,15…to be taken so that the sum is 406 is
[Kerala (Engg.) 2002]
Options:
A) 5
B) 10
C) 12
D) 14
Show Answer
Answer:
Correct Answer: D
Solution:
$ S=\frac{n}{2}[2a+(n-1)d] $
Þ $ 406=\frac{n}{2}[ 6+(n-1)4 ] $
Þ $ 812=n,[6+4n-4] $
Þ $ 812=2n+4n^{2} $
Þ $ 406=2n^{2}+n $
Þ $ 2n^{2}+n-406=0 $
Þ
$ \Rightarrow $ $ =\frac{-1\pm \sqrt{3249}}{4} $ $ \frac{3}{7} $ Taking (+) sign, $ (a-d)a=24 $ .
BETA
