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.

Asked by Pragya Singh | 2 years ago |  140

Solution :-

Let a be the first term and d be the common difference of the A.P.

The sum of the first p terms of the A.P. is given by the equation

$$s_p=\dfrac{p}{2}[(2a+(p-1)d]$$ and the first term q by

$$s_q=\dfrac{q}{2}[(2a+(q-1)d]$$

According to the conditions given in the question,

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

$$d=\dfrac{-2a}{p+q-1}$$

Therefore, sum of first p+q terms of the A.P. is given by the equation

$$s_{p+q}=\dfrac{p+q}{2}[2a+(p+q-1)d]$$

Substituting the value of d in the equation we get,

$$s_{p+q}=\dfrac{p+q}{2}[2a+(p+q-1)\dfrac{-2a}{(p+q-1)}]$$

$$\dfrac{p+q}{2}[2a-2a]=0$$

Therefore, 0 is the sum of the first (p+q) terms of the A.P.

Answered by Abhisek | 2 years ago

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