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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)
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Solution
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.
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