Answer :
Answer:
a = [tex]\frac{12}{7}[/tex], r = [tex]\frac{5}{7}[/tex]
Step-by-step explanation:
The sum to infinity of a geometric progression is
[tex]\frac{a}{1-r}[/tex] ; | r | < 1
Thus for first progression
[tex]\frac{a}{1-r}[/tex] = 6 ( multiply both sides by (1 - r) )
a = 6(1 - r) → (1)
Second progression
[tex]\frac{2a}{1-r^2}[/tex] = 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 = [tex]\frac{5}{7}[/tex]
Substitute this value into (1)
a = 6(1 - [tex]\frac{5}{7}[/tex] ) = 6 × [tex]\frac{2}{7}[/tex] = [tex]\frac{12}{7}[/tex]