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.