In an arithmetic progression, if $\mathrm{S}_{40}=1030$ and $\mathrm{S}_{12}=57$, then $\mathrm{S}_{30}-\mathrm{S}_{10}$ is equal to :
Solution
<p>Let a & d are first term and common diff of an AP.</p>
<p>$$\begin{aligned}
& \mathrm{S}_{40}=\frac{40}{2}[2 \mathrm{a}+39 \mathrm{~d}]=1030 \quad\text{..... (1)}\\
& \mathrm{~S}_{12}=\frac{12}{2}[2 \mathrm{a}+11 \mathrm{~d}]=57 \quad\text{..... (2)}
\end{aligned}$$</p>
<p>by (1) & (2)</p>
<p>$$\begin{aligned}
& \mathrm{a}=-\frac{7}{2} \quad \mathrm{~d}=\frac{3}{2} \\
& \therefore \mathrm{~S}_{30}-\mathrm{S}_{10}=\frac{30}{2}[2 \mathrm{a}+29 \mathrm{~d}]-\frac{10}{2}[2 \mathrm{a}+9 \mathrm{~d}] \\
& =20 \mathrm{a}-390 \mathrm{~d} \\
& =515
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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