Question

If 8th term of an AP is 15, the Sum of the 15 its term is​

Answer 1

${ \huge{ \boxed{ \bf{\underline{ \red{Answer}}}}}} : -$

$\begin{gathered} \bf \: S_n = \frac{n}{2} \{2a + (n - 1)d \} \: \: \: \: \: and \\ \\ \bf \: a_n = a + (n - 1)d \\ \\ { \bf \: here} \\ \bf a_8 = 15 \\ \sf \: so \\ \bf \: { \: a + (n - 1)d = 15 \: \: \: \: \: }\\ \bf \implies\: a + (8 - 1)d = 15 \\ \bf \implies\: { a + 7d = 15 \: \: \: - - - (1)} \: \\ \bf \: now \\ \\ \bf \: S_{15} = \frac{15}{2} \{2a + (15 - 1)d \} \\ \bf \implies\: S_n = \frac{15}{2} (2a + 14d) \\ \bf \implies\: S_n = \frac{15}{2} \times 2 {(a + 7d)} \\ \bf \implies\: S _n = 15 \times {15 }\\ \bf \implies\: S_n = 225\end{gathered}$

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Answer 2

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