The 10^th, 16^th and 22^nd terms of a G.P. are x, y and z respectively. Show that x, y and z are in G.P.

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

By formula,

⇒ a_n = ar^{n - 1}

Given,

The 10^th, 16^th and 22^nd terms of a G.P. are x, y and z respectively.

⇒ a₁₀ = x

⇒ x = ar^{10 - 1}

⇒ x = ar⁹ ………..(1)

⇒ a₁₆ = y

⇒ y = ar^{16 - 1}

⇒ y = ar¹⁵ ………..(2)

⇒ a₂₂ = z

⇒ z = ar^{22 - 1}

⇒ z = ar²¹ ………..(3)

Dividing equation (2) by (1), we get :

$\Rightarrow \dfrac{y}{x} = \dfrac{ar^{15}}{ar^9} \\[1em] \Rightarrow \dfrac{y}{x} = \dfrac{r^{15}}{r^9} \\[1em] \Rightarrow \dfrac{y}{x} = r^{15 - 9} \\[1em] \Rightarrow \dfrac{y}{x} = r^6.$

Dividing equation (3) by (2), we get :

$\Rightarrow \dfrac{z}{y} = \dfrac{ar^{21}}{ar^{15}} \\[1em] \Rightarrow \dfrac{z}{y} = \dfrac{r^{21}}{r^{15}} \\[1em] \Rightarrow \dfrac{z}{y} = r^{21 - 15} \\[1em] \Rightarrow \dfrac{z}{y} = r^6.$

Since, $\dfrac{y}{x} = \dfrac{z}{y}$ = r⁶.

Hence, proved that x, y and z are in G.P.