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STD 11 - 8. sequence and series — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 8. sequence and series4 Marks
MCQ
If $a_n=\frac{-2}{4 n^2-16 n+15}$, then $a_1+a_2+\ldots \ldots+a_{25}$ is equal to :
  • A
    $\frac{51}{144}$
  • B
    $\frac{49}{138}$
  • ✓
    $\frac{50}{141}$
  • D
    $\frac{52}{147}$
✓

Answer

Correct option: C.
$\frac{50}{141}$
c
If $a_n=\frac{-2}{4 n^2-16 n+15}$ then $a_1+a_2+\ldots \ldots \ldots a_{25}$

$\sum \limits_{n=1}^{25} a_n=\sum \frac{-2}{4 n^2-16 n+15}$

$=\sum \frac{-2}{4 n^2-6 n-10 n+15}$

$=\sum \frac{-2}{2 n(2 n-3)-5(2 n-3)}$

$=\sum \frac{-2}{(2 n-3)(2 n-5)}$

$=\sum \frac{1}{2 n-3}-\frac{1}{2 n-5}$

$=\frac{1}{47}-\frac{1}{(-3)}$

$=\frac{50}{141}$

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