NEED HELP PLEASE! - find the least number of terms of the geometric series 108, 36, 12,... that must be added together so that the sum differs from the sum to infinity by less than 0.05

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embizze | High School Teacher | (Level 1) Educator Emeritus

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Given the geometric series 108,36,12,...

The infinite sum exists since `r=1/3<1` . The infinite sum is found by `S=a_1/(1-r)=108/(1-1/3)=162`

We want to find the smallest finite n such that `162-S_n<=.05`

Now the finite sum of a geometric series is found by `S_n=a_n((1-r^n)/(1-r))`

So `162-108((1-(1/3)^n)/(1-1/3))<=.05`

`162-108((1-(1/3)^n)/(2/3))<=.05`

`162-162(1-(1/3)^n)<=.05` Subtract 162 and divide by -162:

`1-(1/3)^n>= .999691358` Subtract 1 and divide by negative 1:

`(1/3)^n<=.000308641975` Take the natural log of both sides

`nln(1/3)<=ln(.000308641975)` Divide by ln(1/3) **<0

`n>=(ln(.000308641975))/(ln(1/3))=7.357762781`

So `n>=8` for integer answers

Check: `162-108((1-(1/3)^8)/(1-1/3))=.024691358<.05` as required.

** If you don't have logarithms, you can use guess and check on the inequality.

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