If S_n=4+11+21+34+50+ to n terms, then (1)/(60)(S_29-S_9) is equal to :

Question

If ${S}_{n}=4+11+21+34+50+\ldots$ to $n$ terms, then $\frac{1}{60}({S}_{29}-{S}_{9})$ is equal to :

Accepted Answer

Given that

$$
\begin{aligned}
& \mathrm{S}_n=4+11+21+24+50+\ldots+\mathrm{T}_n \\
& \mathrm{~S}_n=~~~~~~~~4+11+21+34++\mathrm{T}_{n-1}+\mathrm{T}_n \
& -\quad-\quad-\quad-\quad-\quad-\quad- \
&  0=4+7+10+13+16+\ldots\left(\mathrm{T}_n-\mathrm{T}_{n-2}\right)-\mathrm{T}_n
\end{aligned}
$$

$$
\begin{aligned}
& \Rightarrow \mathrm{T}_n=4+7+10+13+16+\ldots . \text { to } n \text { terms } \\
& \Rightarrow \mathrm{T}_n=\frac{n}{2}[2 \times 4+(n-1) 3] \\
& \mathrm{T}_n=\frac{3}{2} n^2+\frac{5}{2} n
\end{aligned}
$$

$$
\begin{aligned}
& \text { So } \mathrm{S}_n=\Sigma \mathrm{T}_n=\frac{3}{2} \Sigma n^2+\frac{5}{2} \Sigma n \\
& \Rightarrow \mathrm{S}_n=\frac{3}{2} \times \frac{n(n+1)(2 n+1)}{6}+\frac{5}{2} \times \frac{n(n+1)}{2}
\end{aligned}
$$

$$
\begin{aligned}
& \Rightarrow S_n=\frac{n(n+1)}{4}(2 n+1+5)=\frac{n(n+1)(n+3)}{2} \\
& \text { Hence, } \frac{1}{60}\left(S_{29}-S_9\right)=\frac{1}{60} \times \frac{1}{2}(29 \times 30 \times 32-9 \times 10 \times 12) \\
& =223
\end{aligned}
$$

Solutions · 1