If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first (p + q) terms

Dear Chanchal,

equate sum of first p and q terms then write d multiplied by its coefficients on one side and a and its coefficients on the other and cancel (p-q) factor, collect all terms one side and you will get expression similar to that of  p+q term's sum which will give you value of sum as 0. 

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A/Q,

  p/2(2a+(p-1)d)=q/2(2a+(q-1)d)

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

=)2ap+p(p-1)d=2aq+q(q-1)d

=)2ap-2aq=d(q(q-1)-p(p-1))

 =)2a(p-q)=d(q^2-p^2+p-q)

after solving you get,  2a=-(p+q-1)d............................(1)

NOW

sum upto (p+q) terms = (p+q)/2(2a+(p+q-1)d)

                                 =(p+q)/2(2a-2a)                                  (from 1)

                                 =0.

best of luck.........................

like my answer?

let the first term and common difference of AP is a and d....

sum of first q terms =sum of first p terms =k

q/2{2a+(q-1)d} = p/2{2a+(p-1)d} = k 

2a = 2k(q+p-1)/pq   and d=-2k/pq

now sum of first p+q terms =Sp+q= p+q/2 {2a +(p+q-1)d}

                                     =(p+q)/2{2k(p+q-1)/pq  - 2k(p+q-1)/pq }

                                    =0

one question to all of you solved this wat will be the sum when p-q=0 that is p=q this is the difficult case. why you neglected this case.

The sum to n terms of an AP is given by the general expression 

We have 

Since p,q are assumed unequal, we have  so that the sum of (p+q)

terms is 0