Question

Let $a_1,a_2$, ____, $a_{30}$ be an A.P., $S=\sum _{i=1}^{30}{a}_i$ and $T=\sum _{i=1}^{15}{a}_{\left(2i-1\right)}$ If $a_5=27$ and $S-2T=75$, then $a_{10}$ is equal to

Answer 1

Answer: 52

$S=\sum _{i=1}^{30}{a}_i=\frac{30}{2}\left[2a_1+29d\right]$
$T=\sum _{i=1}^{15}{a}_{\left(2i-1\right)}=\frac{15}{2}\left[2a_1+28d\right]$
Since, $S-27=75$
$\Rightarrow 30a_1+435d-30a_1-420d=75$
$\Rightarrow d=5$
Also, $a_s=27$
$\Rightarrow a_1+4d=27$
$\Rightarrow a_1=7$
Hence, $a_{10}=a_1+9d=7+9\times 5=52$