# If the sum to infinity of the series: x + (2x/3) + (4x/9) + ... is fifteen, what is x?

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Looking the terms of the series given to us we can see that there is a common ratio between consecutive terms equal to (2x/3)/x = (4x/9)/(2x/3) = 2/3. The series is a geometric progression with the first term equal to x and a common ratio of 2/3 which is less than 1.

The sum of infinite terms of a GP with a common ratio less than 1 can be calculated. It is given by a/(1 – r), where a is the first term and r is the common ratio.

Here we have x/(1 - 2/3) = 15

=> x/(1/3) = 15

=> x = 15*(1/3)

=> x = 5

**The required value of x = 5**

We notice that we could re-write the sum:

x + x*(2/3) + x*(2/3)^2 + ... + x*(2/3)^n, n-> infinite

We'll factorize by x:

x[1 + 2/3 + (2/3)^2 + ... +(2/3)^n] = 15

x*lim[1 + 2/3 + (2/3)^2 + ... +(2/3)^n] = 15

x*lim [1-(2/3)^(n+1)]/(1-2/3) = 15

x*lim 3*[1-(2/3)^(n+1)] = 15

3x*lim[1-(2/3)^(n+1)] = 3x - lim (2/3)^(n+1) = 15

3x - 0 = 15

3x = 15

x = 5

**The value of x, for the sum of the infinite number of terms of the series is 15, is x = 5.**