Sequences And Series Question 9
Question 9 - 30 January - Shift 1
If $a_n=\frac{-2}{4 n^{2}-16 n+15}$, then $a_1+a_2+\ldots \ldots+a _{25}$ is equal to :
(1) $\frac{51}{144}$
(2) $\frac{49}{138}$
(3) $\frac{50}{141}$
(4) $\frac{52}{147}$
Show Answer
Answer: (3)
Solution:
Formula: General Term of a H.P.
Option (3)
If $a_n=\frac{-2}{4 n^{2}-16 n+15}$ then $a_1+a_2+\ldots \ldots . . a _{25}$
$\Rightarrow \quad \sum _{n=1}^{25} a_n=\sum \frac{-2}{4 n^{2}-16 n+15}$
$=\sum \frac{-2}{4 n^{2}-6 n-10 n+15}$
$=\sum \frac{-2}{2 n(2 n-3)-5(2 n-3)}$
$=\sum \frac{-2}{(2 n-3)(2 n-5)}$
$=\sum \frac{1}{2 n-3}-\frac{1}{2 n-5}$
$=\frac{1}{47}-\frac{1}{(-3)}$
$=\frac{50}{141}$
