Question

Let $a_{1}, a_{2}$, ____, $a_{30}$ be an A.P., $S=\displaystyle\sum_{i=1}^{30} a_{i}$ and $T= \displaystyle\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

$S=\displaystyle\sum_{i=1}^{30} a_{i} =\frac{30}{2} \left[2a_{1}+29d\right]$
$T=\displaystyle\sum_{i=1}^{15} a_{\left(2i-1\right)}=\frac{15}{2}\left[2a_{1}+28d\right]$
Since, $S - 27= 75$
$\Rightarrow 30\,a_{1}+435d-30a_{1}-420d=75$
$\Rightarrow d=5$
Also, $a_{s}=27 $
$\Rightarrow a_{1}+4d=27$
$\Rightarrow a_{1}=7 $
Hence, $a_{10}=a_{1}+9d=7+9\times5=52$