If the sum of first 11 terms of an A.P.
Question:

If the sum of first 11 terms of an A.P., $a_{1}, a_{2}, a_{3}, \ldots$ is 0 $\left(a_{1} \neq 0\right)$, then the sum of the A.P., $a_{1}, a_{3}, a_{5}, \ldots, a_{23}$ is $k a_{1}$, where $k$ is equal to :

1. (1) $-\frac{121}{10}$

2. (2) $\frac{121}{10}$

3. (3) $\frac{72}{5}$

4. (4) $-\frac{72}{5}$

Correct Option: , 4

Solution:

Let common difference be $d$.

$\because S_{11}=0 \quad \therefore \frac{11}{2}\left\{2 a_{1}+10 \cdot d\right\}=0$

$\Rightarrow a_{1}+5 d=0 \Rightarrow d=-\frac{a_{1}}{5}$

Now, $S=a_{1}+a_{3}+a_{5}+\ldots+a_{23}$

$=a_{1}+\left(a_{1}+2 d\right)+\left(a_{1}+4 d\right)+\ldots . .+\left(a_{1}+22 d\right)$

$=12 a_{1}+2 d \frac{11 \times 12}{2}=12\left[a_{1}+11 \cdot\left(-\frac{a_{1}}{5}\right)\right]$ (From (i))

$=12 \times\left(-\frac{6}{5}\right) a_{1}=-\frac{72}{5} a_{1}$