If 3rd and 6th terms of a geometric progression are 3 and 1/9 respectively, then what is the sum of infinite terms of that geometric progression?

If 3rd and 6th terms of a geometric progression are 3 and 1/9 respectively, then what is the sum of infinite terms of that geometric progression? Correct Answer 40.5

GIVEN:

3^rd and 6^th terms of a geometric progression are 3 and 1/9 respectively.

CONCEPT:

Geometric progression formulas for calculating n^th term and sum of infinite terms.

FORMULA USED:

n^th term of a geometric progression = ar^{(n - 1)}

Sum of infinite terms of a geometric progression:

⇒ a / (r - 1) if r > 1

⇒ a / (1 - r) if r < 1

Where

a = first term, r = common ratio, n = number of terms

CALCULATION:

3^rd and 6^th terms of a geometric progression are 3 and 1 / 9 respectively.

Since the value of 6^th term is less than 3^rd term so r < 1.

Using the formula:

ar^{(3 - 1)} = 3

⇒ ar² = 3

And

ar^{(6 - 1)} = 1 / 9

⇒ ar⁵ = 1 / 9

From both the equations:

⇒ r³ = 1 / 27

⇒ r = 1 / 3

And a = 27