Question

A geometric progression has first term a ,common ratior and sum to infinity 6. A second

geometric progression has first term 2a. common ratio r^2and sum to infinity 7. Find the values
of a and r

Answer 1

a =
`(12)/(7)`, r =
`(5)/(7)`

Explanation:

The sum to infinity of a geometric progression is

`(a)/(1-r)` ; | r | < 1

Thus for first progression

`(a)/(1-r)` = 6 ( multiply both sides by (1 - r) )

a = 6(1 - r) → (1)

Second progression

`(2a)/(1-r^2)` = 7 ← multiply both sides by (1 - r² )

2a = 7(1 - r² ) = 7(1 - r)(1 + r) ← difference of squares

2a = 7(1 - r)(1 + r) → (2)

Substitute a = 6(1 - r) into (2)

2(6(1 - r) = 7(1 - r)(1 + r)

12(1 - r) = 7(1 - r)(1 + r) ← divide both sides by (1 - r)

12 = 7(1 + r) = 7 + 7r ( subtract 7 from both sides )

5 = 7r ( divide both sides by 7 )

r =
`(5)/(7)`

Substitute this value into (1)

a = 6(1 -
`(5)/(7)` ) = 6 ×
`(2)/(7)` =
`(12)/(7)`