Let a_{1} and d_{1} be the first term and the common difference of the first arithmetic

progression respectively and a_{2} and d_{2} be the first term and the common difference of the second arithmetic progression respectively.

According to the conditions given in the question,

Sum of n terms of first A.P / Sum of n terms of second A.P.

\( \dfrac{5n+4}{9n+6}\)

\(\dfrac{\dfrac{n}{2}[2a_1+(n-1)d_1]}{\dfrac{n}{2}[2a_2+(n-1)d_2]}= \dfrac{5n+4}{9n+6}\)

\(\dfrac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}= \dfrac{5n+4}{9n+6} \)

Substitute n = 35 in the equation.

\( \dfrac{2a_1+34d_1}{2a_2+34d_2}= \dfrac{5(35)+4}{9(35)+6}\)

= \( \dfrac{a_1+17d_1}{a_2+17d_2}= \dfrac{179}{321}\)

We also know that 18^{th} term of first A.P. /18^{th} term of second A.P.

\( \dfrac{a_1+17d_1}{a_2+17d_2}\)

Then,

18^{th} term of first A.P. /18^{th} term of second A.P.= \( \dfrac{179}{321}\)

Therefore, 179 : 321 is the ratio of 8^{th} term of both the arithmetic progressions.

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