Hello!

If we denote the common ratio as `r,` then the second term is `25r` and the third is `25r^2.`

The sum of the first three terms is `25(1+r+r^2)` and it is given to be `61.`

So we obtained the equation for `r,`

`r^2+r+1=61/25,` or `r^2+r-36/25=0.`

We can solve for...

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Hello!

If we denote the common ratio as `r,` then the second term is `25r` and the third is `25r^2.`

The sum of the first three terms is `25(1+r+r^2)` and it is given to be `61.`

So we obtained the equation for `r,`

`r^2+r+1=61/25,` or `r^2+r-36/25=0.`

We can solve for r using the quadratic formula. `(-1 +- sqrt(1^2 - 4*1*(-36/25)))/(2(1))`

Which can be simplified to `(-1 +- sqrt(169/25))/2`

So the answer for the first question is **-9/5** and **4/5**.

For a geometric progression to have a finite sum of all its terms it is necessary and sufficient to have a common ratio with the absolute value less then 1. So only `r_2=4/5` is suitable. The sum of all terms is `25/(1-r_2)=(25)/(1/5)=` **125**. This is the answer for the second question.