Question

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

Answer 1

Answer:

a = $\frac{12}{7}$, r = $\frac{5}{7}$

Step-by-step explanation:

The sum to infinity of a geometric progression is

$\frac{a}{1-r}$ ; | r | < 1

Thus for first progression

$\frac{a}{1-r}$ = 6 ( multiply both sides by (1 - r) )

a = 6(1 - r) → (1)

Second progression

$\frac{2a}{1-r^2}$ = 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 = $\frac{5}{7}$

Substitute this value into (1)

a = 6(1 - $\frac{5}{7}$ ) = 6 × $\frac{2}{7}$ = $\frac{12}{7}$