The 4th term of a G.P. is 16 and the 7th term is 128. Find the first term and common ratio of the series.

Let first term be a and common ratio be r.

We know that,

nth term of a G.P. is given by,

T_n = ar^{n - 1}

Given,

4th term of a G.P. is 16.

⇒ ar^{4 - 1} = 16

⇒ ar³ = 16 ……(1)

Given,

7th term is 128.

⇒ ar^{7 - 1} = 128

⇒ ar⁶ = 128 ……(2)

Dividing Equation (2) by Equation (1) :

$\Rightarrow \dfrac{ar^6}{ar^3} = \dfrac{128}{16} \\[1em] \Rightarrow r^{6 - 3} = 8 \\[1em] \Rightarrow r^3 = 8 \\[1em] \Rightarrow r = \sqrt[3]8 \\[1em] \Rightarrow r = 2.$

Substituting r = 2 in Equation 1, we get :

⇒ a(2)³ = 16

⇒ a(8) = 16

⇒ a = $\dfrac{16}{8}$

⇒ a = 2.

Hence, a = 2 and r = 2.