JEE PYQ: Sequence And Series Question 20
Question 20 - 2020 (02 Sep Shift 2)
If the sum of first 11 terms of an A.P., $a_1, a_2, a_3, \ldots$ is 0 $(a_1 \neq 0)$, then the sum of the A.P., $a_1, a_3, a_5, \ldots, a_{23}$ is $ka_1$, where $k$ is equal to:
(1) $-\frac{121}{10}$
(2) $\frac{121}{10}$
(3) $\frac{72}{5}$
(4) $-\frac{72}{5}$
Show Answer
Answer: (4)
Solution
Let common difference be $d$.
$\therefore S_{11} = 0 \therefore \frac{11}{2}{2a_1 + 10 \cdot d} = 0$
$\Rightarrow a_1 + 5d = 0 \Rightarrow d = -\frac{a_1}{5}$
Now, $S = a_1 + a_3 + a_5 + \ldots + a_{23}$
$= a_1 + (a_1 + 2d) + (a_1 + 4d) + \ldots + (a_1 + 22d)$
$= 12a_1 + 2d \cdot \frac{11 \times 12}{2} = 12\left[a_1 + 11\left(-\frac{6}{5}\right)\right]$
$= 12 \times \left(-\frac{6}{5}\right)a_1 = -\frac{72}{5}a_1$
BETA
