Question

If the sum of first p terms of an A.P. is equal to the sum of first q terms then show that the sum of its first (p + q) terms is zero. (p ≠ q)

Answer 1

Given that `S_p=S_n`

We know, 

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

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

`p/2(2a+(p-1)d)=q/2(2a+(q-1)d)`

`p(2a+(p-1)d)=q(2a+(q-1)d)`

`2ap+(p-1)dp=2aq+(q-1)dq`

`2ap-2aq+(p-1)dp-(q-1)dq=0`

`2a(p-q)+d(p^2-p-q^2+q)=0`

`2a(p-q)+d[p^2-q^2-1(p-q)]=0`

`2a(p-q)+d[(p+q)(p-q)-1(p-q)]=0`

`2a(p-q)+d(p-q)[p+q-1]=0`

p ≠ q

p - q ≠ 0

Dividing throughout by (p - q),since p ≠ q.

`2a+(p+q-1)d=0` ... (i)

Using eq (i)

`S_(p+q)=(p+q)/2(2a+(p+q-1)d)`

`S_(p+q)=(p+q)/2xx0`

`S_(p+q)=0`

Hence, the sum of its first (p + q) terms is zero.