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

a_{n}=a+(n-1)d

According to the conditions given in the question,

p^{th} term can be written as a_{p}=a+(p-1)d = \( \dfrac{1}{p}\)

q^{th} term can be written as a_{q}=a+(q-1)d = \( \dfrac{1}{q}\)

Subtract a_{q} from a_{p}

(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 agoConstruct a quadratic in x such that A.M. of its roots is A and G.M. is G.

Find the geometric means of the following pairs of numbers:

**(i) **2 and 8

**(ii) **a^{3}b and ab^{3}

**(iii) **–8 and –2

Insert 5 geometric means between \( \dfrac{32}{9}\) and \( \dfrac{81}{2}\).