Question

If the 4\(^{th}\), 10\(^{th}\) and 16\(^{th}\) terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.

Answer 1

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

According to the given condition,

\(a_4=a\space r^3\) = x … (1)

\(a_{10}= a\space r^9\) = y … (2)

\(a_{16}=a\space r^{15}\) = z … (3)

Dividing (2) by (1), we obtain

\(\frac{y}{x}=\frac{ar^9}{ar^3}\) ⇒ \(\frac{y}{x}=r^6\)

Dividing (3) by (2), we obtain

\(\frac{z}{y}=\frac{ar^{15}}{ar^9}\) ⇒ \(\frac{z}{y}=r^6\)

∴\(\frac{y}{x}=\frac{z}{y}\)

Thus, x, y, z are in G. P.