Question

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

Answer 1

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

$S_m$ = $S_n$

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

2a (m - n) + {m (m - 1) - n (n - 1)d} = 0

2a (m - n) + {($m^2$ - $n^2$) - (m - n) d} = 0

(m - n)[2a + (m + n - 1)d] = 0

2a + (m + n - 1) = 0

2a + (m + n -1)d = 0 [Since , m-n $\ne$ 0] ..(i)

Now , $S_{m+n}$ = ($\frac{m+n}{2}$) {2a + (m + n - 1)d}

$S_{m+n}$ = ($\frac{m+n}{2}$) $\times$ 0 = 0[Using equation (i)]