Question

Sum of first P terms of an AP is 0 , if 'a' is the first term , prove that the sum of next q terms is -aq(p+q)/p-1, P is not equal to 0

Answer 1

Let
`d` be the common difference between terms. Then the AP is


`\left\{\underbrace{a,a+d,a+2d,\ldots,a+(p-1)d}_{\text{first }p\text{ terms}},\underbrace{a+pd,a+(p+1)d,\ldots,a+(p+q+1)d}_{\text{next }q\text{ terms}}\right\}`

Because the first
`p` terms add to zero, you have


`\displaystyle\sum_(n=1)^p (a+(n-1)d)=(a-d)\sum_(n=1)^p1+d\sum_(n=1)^pn=(a-d)p+d\frac{p(p+1)}2=0`

Solving for
`d` yields
`d=(2a)/(1-p)`.

Now, the sum of the next
`q` terms is


`\displaystyle\sum_(n=p+1)^(p+q)(a+(n-1)d)=\sum_(n=p+1)^(p+q)\left(a+(n-1)(2a)/(1-p)\right)`

`=\displaystyle\left(a-(2a)/(1-p)\right)\sum_(n=p+1)^(p+q)1+(2a)/(1-p)\sum_(n=p+1)^(p+q)n`

`=\displaystyle\left(a-(2a)/(1-p)\right)q+(2a)/(1-p)\left(\frac{q(2p+q+1)}2\right)`

`=\displaystyle(aq(p+q))/(1-p)`

`=\displaystyle-(aq(p+q))/(p-1)`

as desired.