Notice that
a + arĀ³ + arā¶ + arā¹ + ... = a + rĀ³ (a + arĀ³ + arā¶ + ...)
but the grouped sum on the right side is the same as the sum on the left side. This means that, if the sum we want to compute
a + arĀ³ + arā¶ + arā¹ + ...
converges to S, then
S = a + rĀ³ S
Solve for S :
S - rĀ³ S = a
(1 - rĀ³) S = a
S = a/(1 - rĀ³)
Similarly, knowing the value of the first sum, we have
a + ar + arĀ² + arĀ³ + ... = a + r (a + ar + arĀ² + ...)
but we know
a + ar + arĀ² + ... = 15
so
15 = a + 15r
and knowing the value of the second sum, we have
a + arĀ² + arā“ + ... = a + rĀ² (a + arĀ² + arā“ + ...)
so
9 = a + 9rĀ²
Solve these two equations for a and r :
15 = a + 15r Ā ā Ā a = 15 - 15r
9 = a + 9rĀ² Ā ā Ā 9 = 15 - 15r + 9rĀ² Ā ā Ā 3rĀ² - 5r + 2 = (3r - 2) (r - 1) = 0
ā Ā r = 2/3 or r = 1
We cannot have r = 1, since that would mean
a + a + a + ... = 15
since the left side is adding infinitely many copies of some constant; such a sum can never converge to a finite number. So it must be that r = 2/3, and it follows that a = 15 - 15 (2/3) = 5.
Then the sum we want has a value of
S = 5/(1 - (2/3))Ā³ = 135/19