Question

Let $a_{1}, a_{2}, a_{3}, \ldots, a_{21}$ be in arithmetic progression. If $\displaystyle\sum_{k=1}^{21} a_{k}=693$, then $\displaystyle\sum_{r=0}^{10} a_{2 r+1}$ equals

Answer 1

Given, $a_{1}+a_{2}+a_{3}+a_{4}+\ldots+a_{21}=693$
$\Rightarrow \frac{21}{2}\left[2 a_{1}+20 d\right]=693$,
where $d$ is the common difference of arithmetic progression
$\Rightarrow 21\left[a_{1}+10 d\right]=693$, so $\left(a_{1}+10 d\right)=33$ ...(1)
Now, $a_{1}+a_{3}+a_{5}+\ldots +a_{19}+a_{21}=\frac{n}{2}\left[2 a_{1}+\left(n_{1}-1\right) 2 d\right]$
$=\frac{11}{2}\left[2 a_{1}+10 \cdot 2 d\right]$
$=11\left[a_{1}+10 d\right]=11 \times 33=363$ [using equation (1)]