If sum of first n terms of an A.P. be equal to sum of its first m terms, (m ne n) , then sum of its

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8. Sequences and Series

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If the sum of first $n$ terms of an $A.P.$ be equal to the sum of its first $m$ terms, $(m \ne n)$, then the sum of its first $(m + n)$ terms will be

A

$0$

B

$n$

C

$m$

D

$m + n$

Solution

(a) As given $\frac{n}{2}\left\{ {2a + (n – 1)d} \right\} = \frac{m}{2}\left\{ {a + (m – 1)d} \right\}$

$ \Rightarrow $$2a(m – n) + d({m^2} – m – {n^2} + n) = 0$

$ \Rightarrow $ $(m – n)\left\{ {2a + d(m + n – 1)} \right\} = 0$

$ \Rightarrow $ $2a + (m + n – 1)d = 0$,$(\because \;m \ne n)$

$\therefore $ ${S_{m + n}} = \frac{{m + n}}{2}\left\{ {2a + (m + n – 1)d} \right\} $

$= \frac{{m + n}}{2}\left\{ 0 \right\} = 0$.

Standard 11

Mathematics

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