Let S_n denote the sum of the first n terms o | vedclass

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Question

$ \mathrm{S}_{10}=390 $

$ \frac{10}{2}[2 \mathrm{a}+(10-1) \mathrm{d}]=390 $

$ \Rightarrow 2 \mathrm{a}+9 \mathrm{~d}=78 $      &

Vedclass · Accepted Answer

$790$

Vedclass · Answer

$ \mathrm{S}_{10}=390 $

$ \frac{10}{2}[2 \mathrm{a}+(10-1) \mathrm{d}]=390 $

$ \Rightarrow 2 \mathrm{a}+9 \mathrm{~d}=78 $               $......(1)$

$ \frac{\mathrm{t}_{10}}{\mathrm{t}_5}=\frac{15}{7} \Rightarrow \frac{\mathrm{a}+9 \mathrm{~d}}{\mathrm{a}+4 \mathrm{~d}}=\frac{15}{7} \Rightarrow 8 \mathrm{a}=3 \mathrm{~d} $                        $......(2)$

$ 	ext { From }(1) \&(2) \quad \mathrm{a}=3 \& \mathrm{~d}=8 $

$ \mathrm{~S}_{15}-\mathrm{S}_5=\frac{15}{2}(6+14 	imes 8)-\frac{5}{2}(6+4 	imes 8) $

$ =\frac{15 	imes 118-5 	imes 38}{2}=790$

Let $S_n$ denote the sum of the first $n$ terms of an arithmetic progression. If $\mathrm{S}_{10}=390$ and the ratio of the tenth and the fifth terms is $15: 7$, then $S_{15}-S_5$ is equal to:

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