Question

Let $a_1,a_2,a_3,\dots ,a_{21}$ be in arithmetic progression. If $\sum _{k=1}^{21}{a}_k=693$, then $\sum _{r=0}^{10}{a}_{2r+1}$ equals

Answer 1

Answer: 363

Given, $a_1+a_2+a_3+a_4+\dots +a_{21}=693$
$\Rightarrow \frac{21}{2}\left[2a_1+20d\right]=693$,
where $d$ is the common difference of arithmetic progression
$\Rightarrow 21\left[a_1+10d\right]=693$, so $\left(a_1+10d\right)=33$ ...(1)
Now, $a_1+a_3+a_5+\dots +a_{19}+a_{21}=\frac{n}{2}\left[2a_1+\left(n_1-1\right)2d\right]$
$=\frac{11}{2}\left[2a_1+10\cdot 2d\right]$
$=11\left[a_1+10d\right]=11\times 33=363$ [using equation (1)]