In an A.P., if pth term is $$\dfrac{1}{q}$$ and qth term is $$\dfrac{1}{p}$$, prove that the sum of first pq terms is $$\dfrac{1}{2}$$ (pq + 1) where p ≠ q.

Asked by Pragya Singh | 1 year ago |  120

##### Solution :-

The general term of an A.P. is given by the equation

an=a+(n-1)d

According to the conditions given in the question,

pth term can be written as ap=a+(p-1)d = $$\dfrac{1}{p}$$

qth term can be written as aq=a+(q-1)d = $$\dfrac{1}{q}$$

Subtract aq from ap

(p-1)d-(q-1)d = $$\dfrac{1}{q}- \dfrac{1}{p}$$

(p-1-q+1)d = $$\dfrac{p-q}{pq}$$

(p-q)d = $$\dfrac{p-q}{pq}$$

$$d= \dfrac{1}{pq}$$

Substitute $$d= \dfrac{1}{pq}$$in a+(p-1)d=$$\dfrac{1}{q}$$

a+(p-1)d$$\dfrac{1}{pq}= \dfrac{1}{q}$$

$$a= \dfrac{1}{q}- \dfrac{1}{q}+ \dfrac{1}{pq}= \dfrac{1}{q}$$

The sum of first n terms of an arithmetic progression is given by the equation

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

Substitute the values of n , a and d in the equation.

$$s_{pq}= \dfrac{pq}{2}[\dfrac{2}{pq}+(pq-1)\dfrac{1}{pq}]$$

$$1+ \dfrac{1}{2}(pq-1)$$

$$\dfrac{1}{2}pq+1- \dfrac{1}{2}$$

$$\dfrac{1}{2}pq+\dfrac{1}{2}$$

$$\dfrac{1}{2}(pq+1)$$

Therefore, $$\dfrac{1}{2}(pq+1)$$ is the sum of first pq terms of the A.P.

Answered by Abhisek | 1 year ago

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