The 4th, 6th and last term of a geometric progression are 10, 40 and 640 respectively. If the common ratio is positive, find the first term, common ratio and the number of terms of the series.

We know that,

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

T_n = ar^{n - 1}

Given,

4th term = 10

⇒ T₄ = ar³

⇒ ar³ = 10 …..(1)

Given,

6th term = 40

⇒ T₆ = ar⁵

⇒ ar⁵ = 40 …..(2)

Given,

Let nth term be last term.

⇒ T_n = ar^{n - 1}

Since, last term = 640

⇒ ar^{n - 1} = 640 …..(3)

Divide Equation 2 by Equation 1:

$\Rightarrow \dfrac{ar^5}{ar^3} = \dfrac{40}{10}$

⇒ r^{5 - 3} = 4

⇒ r² = 4

⇒ r = 2 (Since, common ratio is positive)

Substitute r = 2 into Equation 1:

⇒ a(2)³ = 10

⇒ a(8) = 10

⇒ a = $\dfrac{10}{8}$

⇒ a = $\dfrac{5}{4}$.

Now, substituting values of a and r in equation (3), we get :

⇒ $\dfrac{5}{4} \times$ 2^{n - 1} = 640

⇒ 2^{n - 1} = $\dfrac{640 × 4}{5}$

⇒ 2^{n - 1} = 128 × 4

⇒ 2^{n - 1} = 512

⇒ 2^{n - 1} = 2⁹

⇒ n - 1 = 9

⇒ n = 9 + 1

⇒ n = 10.

Hence, a = $\dfrac{5}{4}$, r = 2 and n = 10.