The 4th, 6th and the 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.

For a geometric progression:

Tₙ = arⁿ⁻¹

Given:

T₄ = ar³ = 10 ...(1)

T₆ = ar⁵ = 40 ...(2)

Divide (2) by (1):

r² = 40/10

r² = 4

Since the common ratio is positive:

r = 2

From (1):

a(2³) = 10

8a = 10

a = 5/4

The last term is 640:

(5/4) × 2ⁿ⁻¹ = 640

2ⁿ⁻¹ = 640 × 4/5

= 512

= 2⁹

Therefore:

n − 1 = 9

n = 10

Answer:

First term = 5/4 Common ratio = 2 Number of terms = 10.

The ratio of the 6th term to the 4th term gives r² directly. After finding r and a, substitute the last term in Tₙ = arⁿ⁻¹ to determine n.