Question

If the nth, (2n)^th, (3n)^th terms of a G.P. are a, b, c respectively then show that b₂ = ac.

Answer 1

Let the first term of the G.P. be x and the common ratio be y.
Now,
n^th term of the G.P.:
t_n = xy^{n – 1} = a …(1)
(2n)^th term of the G.P.:
t_2n = xy^{2n – 1} = b …(2)
(3n)^th term of the G.P.:
t_3n = xy^{3n – 1} = c …(3)
On multiplying equations (1) and (3), we get:

$$\begin{gathered} \left(x{y}^{n-1}\right)\left(x{y}^{3n-1}\right)=ac \\ {x}^2{y}^{n-1+3n-1}=ac \\ {x}^2{y}^{4n-2}=ac \\ {x}^2{y}^{2(2n-1)}=ac \\ {\left(x{y}^{(2n-1)}\right)}^2=ac \\ \mathrm{From}\mathrm{equation}\left(2\right),\mathrm{we}\mathrm{get}: \\ {b}^2=ac \end{gathered}$$