If and , then what is the sum of the geometric series ?

Notice that

a + ar³ + ar⁶ + ar⁹ + ... = a + r³ (a + ar³ + ar⁶ + ...)

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 + ar³ + ar⁶ + ar⁹ + ...

converges to S, then

S = a + r³ S

Solve for S :

S - r³ S = a

(1 - r³) S = a

S = a/(1 - r³)

Similarly, knowing the value of the first sum, we have

a + ar + ar² + ar³ + ... = a + r (a + ar + ar² + ...)

but we know

a + ar + ar² + ... = 15

so

15 = a + 15r

and knowing the value of the second sum, we have

a + ar² + ar⁴ + ... = a + r² (a + ar² + ar⁴ + ...)

so

9 = a + 9r²

Solve these two equations for a and r :

15 = a + 15r   ⇒   a = 15 - 15r

9 = a + 9r²   ⇒   9 = 15 - 15r + 9r²   ⇒   3r² - 5r + 2 = (3r - 2) (r - 1) = 0

⇒   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))³ = 135/19

37 = ending with 7

unit's digit in the cube..3

56 = ending with 6

unit's digit in the cube..2

133 = ending with

unit's digit in the cube..7

hope it : ( : )