JEE MAIN - Mathematics (2024 - 30th January Morning Shift - No. 19)
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 :
405
390
410
395
Explanation
$$\begin{aligned} &\begin{aligned} & \mathrm{S}_{20}=\frac{20}{2}[2 \mathrm{a}+19 \mathrm{~d}]=790 \\ & 2 \mathrm{a}+19 \mathrm{~d}=79 \quad \text{.... (1)}\\ & \mathrm{~S}_{10}=\frac{10}{2}[2 \mathrm{a}+9 \mathrm{~d}]=145 \\ & 2 \mathrm{a}+9 \mathrm{~d}=29 \quad \text{.... (2)} \end{aligned}\\ &\text { From (1) and (2) } a=-8, d=5 \end{aligned}$$
$$\begin{aligned} & S_{15}-S_5=\frac{15}{2}[2 a+14 d]-\frac{5}{2}[2 a+4 d] \\ & =\frac{15}{2}[-16+70]-\frac{5}{2}[-16+20] \\ & =405-10 \\ & =395 \end{aligned}$$
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