Let’s take a and d to be the first term and the common difference of the A.P. respectively.

So, we have

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

Therefore,

\( s_1=\dfrac{n}{2}[2a+(n-1)d]\)

\( s_2=\dfrac{2n}{2}[2a+(n-1)d]\)

= \( n[2a+(n-1)d]\)

\( s_n=\dfrac{3n}{2}[2a+(n-1)d]\)

Subtract S_{1 }from S_{2} .

\( S_2-S_1=S_2\)

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

= \( n[\dfrac{4a+4nd-2d-2a-nd+d}{2}]\)

= \( n[\dfrac{2a+3nd-d}{2}]\)

= \( \dfrac{n}{2}[2a+(3n-1)d]\)

Therefore,

S_{3}= \( \dfrac{3n}{2}[2a+(3n-1)d]\)

= 3(S_{2}-S_{1})

Therefore, S_{3}= 3(S_{2}-S_{1}) is proved

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