Let S_n denote the sum of first n terms of an arithmetic progression.

Question

Let $S_n$ denote the sum of first $n$ terms of an arithmetic progression. If $S_{20}=790$ and $S_{10}=145$, then $$\mathrm{S}_{15}-\mathrm{S}_5$$ is :

Accepted Answer

$$\begin{aligned}
&\begin{aligned}
& \mathrm{S}_{20}=\frac{20}{2}[2 \mathrm{a}+19 \mathrm{~d}]=790 \
& 2 \mathrm{a}+19 \mathrm{~d}=79 \quad \text{.... (1)}\
& \mathrm{~S}_{10}=\frac{10}{2}[2 \mathrm{a}+9 \mathrm{~d}]=145 \
& 2 \mathrm{a}+9 \mathrm{~d}=29 \quad \text{.... (2)}
\end{aligned}\
\&\text { From (1) and (2) } a=-8, d=5
\end{aligned}$$

$$\begin{aligned}
& S_{15}-S_5=\frac{15}{2}[2 a+14 d]-\frac{5}{2}[2 a+4 d] \
& =\frac{15}{2}[-16+70]-\frac{5}{2}[-16+20] \
& =405-10 \
& =395
\end{aligned}$$

Solutions · 1