Question

If in an A.P., the sum of first m terms is n and the sum of its first n terms is m, then prove that the sum of its first (m + n) terms is –(m + n).

Answer 1

Let 1^st term of series be a and common difference be d, then

S_m = n   ...(Given)

⇒ `m/n [2a + (m - 1)d] = n`

⇒ m [2a + (m – 1) d] = 2n   ...(i)

And S_n = m   ...(Given)

⇒ `n/2 [2a + (n - 1) d] = m`

⇒ n [2a + (n – 1) d] = 2m   ...(ii)

On subtracting,

2 (n – m) = 2a (m – n) + d [m² – n² – (m – n)]

⇒ 2 (n – m) = 2a (m – n) + d [(m – n)] [– (m – n) – (m – n)]

⇒ 2 (n – m) = (m – n) [2a + d (m + n – 1)]

⇒ – 2 = 2a + d (m + n – 1)

Now, `S_(m + n) = (m + n)/2 [2a + (m + n - 1) d]`

= `(m + n)/2 (-2)`

= – (m + n)

Hence proved.