Question

If the sum of first $11$ terms of an A.P. $a_1,a_2,a_3.......$is $0(a_1\ne 0)$ then the sum of the A.P. $a_1,a_3,a_5........a_{23}$ is $k{a}_1$, where $k$ is equal to

• A. $-\frac{121}{10}$
• B. $-\frac{72}{5}$
• C. $\frac{72}{5}$
• D. $\frac{121}{10}$

Answer 1

The correct option is B

$-\frac{72}{5}$

Explanation for the correct option.

Step 1: Solve for the relation between $a$ and $d$.

Given: The sum of first $11$ terms of an A.P. is $0$, $\Rightarrow \sum _{k=1}^{11}{a}_k=0$

We know that sum of first $n$ terms of A.P. is given by, $S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$

$$\begin{gathered} \Rightarrow \frac{11}{2}\left[2a+10d\right]=0 \\ \Rightarrow 11\left[a+5d\right]=0 \\ \Rightarrow 11a+55d=0 \\ \Rightarrow a+5d=0 \\ \Rightarrow d=-\frac{a}{5} \end{gathered}$$

Step 2: Solve for the required value.

Given: The sum of the A.P. $a_1,a_3,a_5........a_{23}$ is $k{a}_1$

$$\begin{gathered} a_1+a_3+a_5........a_{23}=k{a}_1 \\ \Rightarrow a+a+2d+a+4d......a+22d=ka\left[\because a_n=a+\left(n-1\right)d\right] \\ \Rightarrow 12a+d(2+4+6........+22)=ka \\ \Rightarrow 12a+2d\left(1+2+3+......11\right)=ka \\ \Rightarrow 12a+2d\left(66\right)=ka\left[\because \sum _{n=1}^{11}n=\frac{n\left(n+1\right)}{2}=\frac{11\times 12}{2}=66\right] \\ \Rightarrow 12a+132d=ka \\ \Rightarrow 12\left(a+11\left(-\frac{a}{5}\right)\right)=ka \\ \Rightarrow 12\left(1-\frac{11}{5}\right)=k \\ \Rightarrow k=12\left(-\frac{6}{5}\right) \\ \Rightarrow k=-\frac{72}{5} \end{gathered}$$

Hence option(B) i.e. $-\frac{72}{5}$ is the correct option.