If the sum of the first 11 terms of an A.P., a_1,a_2,a_3, is 0 (a_10), then the sum of the…

Question

If the sum of the first 11 terms of an A.P., $a_1,a_2,a_3,\ldots$ is $0$ ($a_1
e0$), then the sum of the A.P. $a_1,a_3,a_5,\ldots,a_{23}$ is $ka_1$, where $k$ is equal to:

Accepted Answer

$S_{11}=0 \Rightarrow a_1+5d=0 \Rightarrow d=-\dfrac{a_1}{5}$.

$$a_1+a_3+\cdots+a_{23} = 12a_1+2d\cdot\frac{11\times12}{2} = 12a_1+132d = 12a_1 - \frac{132a_1}{5} = -\frac{72}{5}a_1.$$

So $k=-\dfrac{72}{5}$.

Solutions · 1