Question

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 GP.

Answer 1

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:

⇒ `y/x​ = (ar^15​)/(ar^9)`

⇒ `y/x ​=(r^15)/(r^9)`

⇒ `y/x​ = r^(15 − 9)`

⇒ `x/y​ = r^6`

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

⇒ `z/y = (ar^21)/(ar^15)`

⇒ `z/y =(r^21​)/(r^15)`

⇒ `z/y​ = r^(21−15)`

⇒ `z/y ​= r^6`

Since, `y/x = z/y​ = r^6`

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