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USE CODE: chait6

```				   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
```

6 years ago

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```										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.

Now you can win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.
```
6 years ago
```										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.........................

```
6 years ago
```										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
```
6 years ago
```										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.
```
6 years ago
```										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
```
6 years ago

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