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 where p ≠ q.

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#### Solution

To show : `S_(p+q)=0`

that is, to show : `(p+q)/2(2a+(p+q-1)d)=0`

Given that `S_p=S_n`

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`

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 proved.

Concept: Sum of First n Terms of an AP

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