# In a geometric sequence, a=13, t3=208, and the sum of all terms of the series is 17,745. How many terms are in the series?

There are 6 terms in the series. We are given a geometric series with a (the first term)=13 and the third term as 208. The sum of the finite series is 17,745. We are asked to find the number of terms in the series.

The sum of a finite geometric series can be found by

`S_n=(a(1-r^n))/(1-r)` where r is the common ratio and n the number of terms. Since we know the sum and we are looking for the number of terms, we must determine the common ratio.

1. We know the first and third terms; the second term is the geometric mean between these. Thus `13/x=x/208 ==> x^2=2704 ==> x=52` . If the second term is 52, the common ratio is 4.

2. Alternatively we have a=13 and `ar^2=208 ==> 13r^2=208 ==> r^2=16 ==> r=4`

Now we substitute the known quantities:

`17,745=(13(1-4^n))/(1-4)`
`(1-4^n)=(-3*17,745)/13 ==> (1-4^n)=-4,095`
`-4^n=-4,096 ==> 4^n=4,096 ==> n=6`

If we did not know powers of 4, we can use logarithms to solve for n.

`4^n=4,096 ==> log_4 4^n=log_4 4,096 ==> n=(log 4,096)/(log 4) =6`

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