# A geometric series has three terms. The sum of the three terms is 42. The third term is 3.2 times the sum of other two. What are the terms?

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`a_1+a_2+a_3=42` **(1)**

`a_3=3.2(a_1+a_2)` **(2)**

Now we put `3.2(a_1+a_2)` instead of `a_3` in first equation.

`a_1+a_2+3.2(a_1+a_2)=42`

`4.2a_1+4.2a_2=42`

Now we use the fact that this is geometric sequence which means that `a_n=a_1r^(n-1)` hence we have

`4.2a_1+4.2a_1r=42`

`a_1(4.2(1+r))=42`

`a_1=42/(4.2(1+r))=10/(1+r)`

Now from (2) we have

`a_1r^2=3.2(a_1+a_1r)` now we put `10/(1+r)` instead of `a_1`

`(10r^2)/(1+r)=32/(1+r)+(32r)/(1+r)`

`(10r^2-32r-32)/(1+r)=0` This is equal to 0 only if numerator is equal to 0.

`10r^2-32r-32=0`

When we solve this equation we get two solutions:

`r_1=-4/5` and `r_2=4`

For `r_1` we have `a_1=10/(1-4/5)=50`, `a_2=50 cdot(-4/5)=-40`, `a_3=-40cdot(-4/5)=32`

For `r_2` we have

`a_1=10/(1+4)=2`, `a_2=2 cdot 4=8`, `a_3=8 cdot 4=32`