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 $-\frac{72}{5}$

$\sum _{k=1}^{11}{a}_k=0\Rightarrow 11a+55d=0$
$a+5d=0$

Now $a_1+a_3+....+a_{23}=k{a}_1$
$12a+d(2+4+6+......+22)=ka$
$12a+2d.66=ka$
$12(a+11d)=ka$
$12(a+11(-\frac{a}{5}))=ka$
$12(1-\frac{11}{5})=k$
$k=-\frac{72}{5}$