If the sum of first m terms of an A.P. is n and sum of first n terms of the same A.P. is m, show that sum of first (m + n) terms of it is -(m + n).

Let a be the first term and d be common difference of the A.P.

Given,

Sum of first m terms of an A.P. is n.

⇒ S_m = n

⇒ $\dfrac{m}{2}[2a + (m - 1)d] = n$

⇒ m[2a + (m - 1)d] = 2n

⇒ 2am + m(m - 1)d = 2n ……….(1)

Sum of first n terms of an A.P. is m.

⇒ S_n = m

⇒ $\dfrac{n}{2}[2a + (n - 1)d] = m$

⇒ n[2a + (n - 1)d] = 2m

⇒ 2an + n(n - 1)d = 2m ……….(2)

Subtracting eq. (2) from (1), we get

$\Rightarrow 2n - 2m = 2am + m(m - 1)d - [2an + n(n - 1)d] \\[1em] \Rightarrow -2m + 2n = 2am - 2an + m(m - 1)d - n(n - 1)d \\[1em] \Rightarrow -2(m - n) = 2a(m - n) + m^2d - md - n^2d + nd \\[1em] \Rightarrow -2(m - n) = 2a(m - n) + d(m^2 - n^2) - d(m - n) \\[1em] \Rightarrow -2(m - n) = 2a(m - n) + d(m - n)(m + n) - d(m - n) \\[1em] \Rightarrow -2 = 2a + d(m + n) - d \\[1em] \Rightarrow 2a + d(m + n - 1) = -2 ………(3)$

S_{m + n} = $\dfrac{m + n}{2}[a + a + (m + n - 1)d]$

= $\dfrac{m + n}{2}[2a + (m + n - 1)d]$

= $\dfrac{m + n}{2} \times -2$ [From (3)]

= -(m + n).

Hence, proved that sum of first (m + n) terms of it is -(m + n).