Let a be the first term and r be the common ratio of the G.P.

According to the given condition,

a_{4} = a r^{3} = x … (1)

a_{10} = a r^{9} = y … (2)

a_{16}^{ }= a r^{15 }= z … (3)

On dividing (2) by (1), we get

\( \dfrac{y}{x}= \dfrac{ar^9}{ar^3}\)

= \( \dfrac{y}{x}=r^6\)

Now, divide equation (3) by (2).

\( \dfrac{z}{y}= \dfrac{ar^{15}}{ar^9}\)

\( \dfrac{z}{y}= r^6\)

\( \dfrac{y}{x}= \dfrac{z}{y}\)

Therefore, it is proved that x, y, z are in G. 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 \( \dfrac{81}{2}\).