Question

the sum of first n terms of an A.P. is zero, show that the sum of next term m terms is -am(m+n)/n-1, a being the first t​

Answer 1

Let the first term of the given AP be a and common difference be d

Given that sum of first n terms of the AP is 0.

$S_n = \frac{n}{2}(2a+[n-1]d) = 0$

$2a + (n-1)d = 0 .....(i)$

Let us find the sum of first (m+n) terms of the given AP.

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

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

From equation (i), we get

$S_{m+n}=\frac{m+n}{2}(md).....(ii)$

Take equation (i) again.

$2a + (n-1)d = 0$

$(n-1)d = -2a$

$d = \frac{-2a}{n-1}$

Substitute the value of 'd' in equation (ii)

We get,

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

$S_{m+n} = \frac{-am(m+n)}{n-1}$

We know that sum of first n terms of the AP is zero. So the sum of next m terms of the AP will infact be equal to sum of first (m+n) terms of the AP

HENCE PROVED