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.

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- Jul 1st 2011, 01:15 AMblackhat123Help 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. - Jul 1st 2011, 02:55 AMsa-ri-ga-maRe: 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. - Jul 1st 2011, 03:08 AMemakarovRe: Help required to solve an Arithmatic progression problem
Note also that the claim holds if p does not equal q.

- Jul 1st 2011, 03:23 AMblackhat123Re: 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]

Please help me - Jul 1st 2011, 03:50 AMAlso sprach ZarathustraRe: Help required to solve an Arithmatic progression problem
$\displaystyle S_p=S_q$

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

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

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

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

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

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

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

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

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

$\displaystyle S_{p+q}=0$ - Jul 1st 2011, 03:51 AMIsomorphismRe: Help required to solve an Arithmatic progression problem
- Jul 1st 2011, 04:50 AMblackhat123Re: 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. - Jul 1st 2011, 06:25 AMArchie MeadeRe: Help required to solve an Arithmatic progression problem
You could also use

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

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

for q>p

Then

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

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

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

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

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

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