# Find a number which, when added to each of 2,6,13, gives three numbers in geometric sequence.

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### 1 Answer

To start, let's assign a variable to the number. Let it be x.

And if adding x to 2, 6 and 13 results to a geometric series, we would have:

`2+x` , `6+x` , and `13+x`

Note that in geometric series, the ratio between two consecutive terms are the same and it is called as a common ratio (r). So,

`r = (6+x)/(2+x)` and `r=(13+x)/(6+x)`

To solve for x, set the two equations equal to each other.

`r=r`

`(6+x)/(2+x)=(13+x)/(6+x)`

Then, cross multiply.

`(6+x)(6+x)=(13+x)(2+x)`

`36+12x+x^2=26+15x+x^2`

To combine like terms, move `x^2` from the left to the right side of the equation.

`36+12x+x^2-x^2=26+15x+x^2-x^2`

`36+12x=26+15x`

Also, move 12x to the right.

`36+12x-12x=26+15x-12x`

`36=26+3x`

Then, isolate x.

`36-26=26-26+3x`

`10=3x`

`10/3=x`

**Hence, the number that is added to 2,6 and 13, which gives three numbers in geometric series, is `10/3` .**