# Math Help - Help required to solve an Arithmatic progression problem

1. ## 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.

2. ## Re: Help required to solve an Arithmatic progression problem

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.

3. ## Re: Help required to solve an Arithmatic progression problem

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

4. ## Re: Help required to solve an Arithmatic progression problem

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]

5. ## Re: Help required to solve an Arithmatic progression problem

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.
$S_p=S_q$

$\frac{(2a_1 +(p-1)d)p}{2}=\frac{(2a_1+(q-1)d)q}{2}$

$p(2a_1 +pd-d)=q(2a_1+qd-d)$

$2a_1p+p^2d-pd =2a_1q+q^2d-qd$

$2a_1p-2a_1q +p^2d -q^2d -pd +qd =0$

$2a_1(p-q)+d(p+q)(p-q)-d(p-q)=0$

$(p-q)(2a_1+(p+q)d-d)=0$

$2a_1+(p+q)d-d=0$

$2a_1+((p+q)-1)d=0$

$\frac{(p+q)(2a_1+((p+q)-1)d)}{2}=0$

$S_{p+q}=0$

6. ## Re: Help required to solve an Arithmatic progression problem

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.

7. ## Re: Help required to solve an Arithmatic progression problem

Ya... agreed.

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

Thanks all for your help.

8. ## Re: Help required to solve an Arithmatic progression problem

You could also use

$a+(a+d)+(a+2d)+....+[a+(p-1)d]$

$=a+(a+d)+...+[a+(p-1)d]+[a+pd]+[a+(p+1)d]+....+[a+(q-1)d]$

for q>p

Then

$[a+pd]+[a+(p+1)d]+....+[a+(q-1)d]=0$

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

$0.5[a+pd+a+(q-1)d]=0$

$2a+(p+q-1)d=0$

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

$S_{p+q}=\frac{2a+(p+q-1)d}{2}(p+q)$