First slide

Arithmetic progression

Question

If the sum of $m$ terms of an A.P is the same as the sum of its $n$ terms, then the sum of its $(m + n)$ term is

Moderate

A

$m n$
B

$- m n$
C

$\frac{1}{m n}$
D

0

Solution

Let a be the first term and d be the common differences of the given A.P, then
$\begin{array}{l} S_{m} = S_{n} \Rightarrow \frac{m}{2} [2 a + (m - 1) d] = \frac{n}{2} [2 a + (n - 1) d] \\ \Rightarrow 2 a (m - n) + {m (m - 1) - n (n - 1)} d = 0 \\ \Rightarrow 2 a (m - n) + \{(m^{2} - n^{2}) - (m - n)\} d = 0 \\ \Rightarrow (m - n) [2 a + (m + n - 1) d] = 0 \\ \Rightarrow 2 a + (m + n - 1) d = 0 [∵ m - n \neq 0] \\ Now, \\ S_{m + n} = \frac{m + n}{2} [2 a + (m + n - 1) d] = \frac{m + n}{2} \times 0 = 0 \end{array}$

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