Thread: Help required to solve an Arithmatic progression problem

The sum of the first p terms of an AP in equal to the sum of the
first q terms.Prove that sum of the first p+q terms of the same
AP is zero.

Sum of p terms S1 = (p/2)[2a + (p-1)d]

Sum of q terms S2 = (q/2)[2a + (q-1)d]

Given S1 = S2. Solve the equations and find 2a =

Then put this value in the sum of p+q terms.

Note also that the claim holds if p does not equal q.

I got 2a+d=0 from the equation S1=S2.But I didn't get how this will help to deduce the sum s(p+q)=0 where s(p+q)=(p+q)/2[2a+(p+q-1)d]

Please help me

Originally Posted by blackhat123

The sum of the first p terms of an AP in equal to the sum of the
first q terms.Prove that sum of the first p+q terms of the same
AP is zero.

Originally Posted by blackhat123

I got 2a+d=0 from the equation S1=S2......

Then ,you have made some mistake in your calculation. Show us your working so that we can comment on it. Also, remember emakarov's note and use it in the proof.

Ya... agreed.

I did some silly mistakes.Now I got it solved.

Thanks all for your help.

You could also use





for q>p

Then



Therefore the average must be zero, which is also the average of the first and last terms of this progression.





This causes the sum of p+q terms to be zero, since