JEE Main 2023 — Sequences and Series Question with Solution

Given,

$$\begin{gathered} S_n=4+11+21+34+............+T_n \\ {S}_n= 4+11+21+34......+T_{n-1}+T_n \end{gathered}$$

Subtracting above equations, we get

$0=4+7+10+13+....-T_n$

$\Rightarrow T_n=4+7+10+13+...$

The above series is in AP.

$\Rightarrow T_n=\frac{n}{2}\left(2\times 4+\left(n-1\right)3\right)$

$\Rightarrow T_n=\frac{n}{2}\left(3n+5\right)$

So,

$S_n=\sum _{}^{}\frac{3n^2+5n}{2}$

$\Rightarrow S_n=\frac{1}{2}\left(\frac{3n(n+1)(2n+1)}{6}+\frac{5n\left(n+1\right)}{2}\right)$

$\Rightarrow S_n=\frac{n\left(n+1\right)}{2}\left(\frac{(2n+1)}{2}+\frac{5}{2}\right)$

$\Rightarrow S_{29}=\frac{29\times 30}{2}\left(\frac{59}{2}+\frac{5}{2}\right)$

$\Rightarrow S_{29}=29\times 15\times 32=13920$

$\Rightarrow S_9=\frac{9\times 10}{2}\left(\frac{19}{2}+\frac{5}{2}\right)$

$\Rightarrow S_9=9\times 5\times 12=540$

$\Rightarrow \frac{S_{29}-S_9}{60}=\frac{13920-540}{60}=223$.

Therefore, the required answer is $223$.