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 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 \( \dfrac{81}{2}\).