Question

If the sum of first m terms of an A.P. is same as sum of its first n terms (m ≠ n), then show that the sum of its first (m + n) terms is zero. 

Answer 1

Given: S_m = S_n

Show: S_(m+n) = 0

Let the A.P. be denoted as

a₁, a₂, a₃, ........ a_n ......

With common difference d.

S_m = `m/2[2a_1 + (m - 1)d]`

S_n = `n/2[2a_1 + (n -1)d]`

Given both are equal

`m/2[2a_1 + (m - 1)d] = n/2[2a_1 + (n - 1)d]`

⇒ `1/2 [2a_1m + (m - 1)md] = 1/2[2a_1n + (n - 1)nd]`

⇒ `1/2 [2a_1m + m^2d - md] - 1/2[2a_1n + n^2d - nd] = 0`

⇒ `1/2 [2a_1m - 2a_1n + m^2d - n^2d - md + nd] = 0`

⇒ `1/2 [2a_1(m - n) + d(m^2 - n^2)-d(m - n)= 0]`

⇒ `1/2 (m - n)[2a_1 + (m + n - 1)d] = 0`

2a₁ = − [m + n − 1] d              ...(i)

The sum of the first (m + n) terms of the given A.P.

S_m+n = `(m + n)/2 [2a_1 + (m + n -1)d]`       ...(ii)

Put (i) in (ii)

⇒ S_m+n = `(m + n)/2[-(m + n - 1)d + (m + n -1)d]`

S_m+n = 0