Question

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

Answer 1

a is the first term .

then, sum of n terms in AP when common difference is d .
$S_n = \frac{n}{2}[2a+(n-1)d]\\\\S_n=0=\frac{n}{2}[2a+(n-1)d]\\\\0=2a+(n-1)d\\\\d=\frac{-2a}{(n-1)} - - - (1)$

now, sum of next m terms :
$S_m = S_{m+n}-S_n\\\\=\frac{m+n}{2}[2a+(m+n-1)d]-\frac{n}{2}[2a+(n-1)d]$

= am + d/2[ m² + n² + 2mn - m - n - n² + n ]
= am + d/2[ m² + 2mn - m ]
now, put d = -2a/(n-1)
then, Sm = am + (-2a)/2(n-1)[m² + 2mn - m ]
= am -am/(n-1) [ m + 2n - 1]
= am [ (n -1) - m - 2n +1 ]/(n -1)
= am [n - 1 - m - 2n + 1 ]/(n -1)
= am [ -n - m ]/(n -1)
= -am(m + n )/(n -1)

hence, Sm = -am(m+n)/(n+1)

Answer 2

Answer :

If d is the common difference of this AP, then n/2 [2a + (n - 1)d] = 0

Thus, n/2 [2a + (n - 1)d] = 0

2a + (n - 1)d = 0

(n - 1)d = -2a

d = -2a/( n - 1)

Now, sum of first m terms,

Sum of m terms = m+n/2 [2a + (m + n - 1)d]

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

= (m + n)[ a + (m + n - 1)(-a)/(n - 1)]

= (m + n)[ a (n - 1) + (m + n - 1)(-a)/(n - 1)]

= (m + n)/(n - 1) [an - a - am - an + a)

=(m + n)(-am)/(n - 1)

= - (am)(m + n)/(n - 1)

Therefore, sum of next m terms = Sum of (m + n) terms - Sum of n terms

But, sum of n terms = 0

Therefore, sum of next m terms = Sum of (m+n) terms

= -(am)(m + n)/ (n - 1)