The sum of first p- terms terms of an A.P. is q and the sum of first q terms is p, find the sum of first (p + q)

Explanation

Let a be the first term and d be the common difference of the given A.P.
Then `S_(p)=S_(q)` ...(Given )
Using formula, `S_(n)=(n)/(2)[2a+(n-1)d]`
`(p)/(2)[2a+(p-1)d]=(q)/(2)[2a+(q-1)d]`
`thereforep[2a+(p-1)d]=q[2a+(q-1)d]` ...(Multiplying both the sides by 2)
`thereforep(2a)+p(p-1)d-q(2a)-q(q-1)d=0`
`thereforep(2a)-q(2a)+[p(p-1)-q(q-1)]d=0`
`therefore2a(p-q)+[p^(2)-p-q^(2)+q]d=0`
`therefore2a(p-q)+[p^(2)-q^(2)-p+q]d=0`
`therefore2a(p-q)+[(p-q)(p+q)-(p-q)]d=0`
`therefore(p-q)[2a+(p+q-1)d]=0`
But, `pneq` ...(Given)
`therefore(p-q)ne0`
`therefore2a+(p+q-1)d=0` ...(1)
Now, `S_(p+q)=(p-q)/(2)[2a+(p+q-1)d]`
`=(p+q)/(2)xx0` ...[From(1)]
`=0`
`thereforeS_(p+q)=0`
`therefore` the sum of first `(p+q)` terms is zero.