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 :
Solution
<p>$$\begin{aligned}
&\begin{aligned}
& \mathrm{S}_{20}=\frac{20}{2}[2 \mathrm{a}+19 \mathrm{~d}]=790 \\
& 2 \mathrm{a}+19 \mathrm{~d}=79 \quad \text{.... (1)}\\
& \mathrm{~S}_{10}=\frac{10}{2}[2 \mathrm{a}+9 \mathrm{~d}]=145 \\
& 2 \mathrm{a}+9 \mathrm{~d}=29 \quad \text{.... (2)}
\end{aligned}\\
&\text { From (1) and (2) } a=-8, d=5
\end{aligned}$$</p>
<p>$$\begin{aligned}
& S_{15}-S_5=\frac{15}{2}[2 a+14 d]-\frac{5}{2}[2 a+4 d] \\
& =\frac{15}{2}[-16+70]-\frac{5}{2}[-16+20] \\
& =405-10 \\
& =395
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
This question is part of PrepWiser's free JEE Main question bank. 209 more solved questions on Sequences and Series are available — start with the harder ones if your accuracy is >70%.