Question

The first term of a finite geometric series is 6 and the last term is 4374. The sum of all the term os 6558. find the common ratio and calculate the number of terms in the series.

Answer 1

A geometric series is written as $ar^n$, where $a$ is the first term of the series and $r$ is the common ratio.

In other words, to compute the next term in the series you have to multiply the previous one by $r$.

Since we know that the first time is 6 (but we don't know the common ratio), the first terms are

$6, 6r, 6r^2, 6r^3, 6r^4, 6r^5, \ldots$.

Let's use the other information, since the last term is $4374 > 6$, we know that $r>1$, otherwise the terms would be bigger and bigger.

The information about the sum tells us that

$\displaystyle \sum_{i=0}^n 6r^i = 6\sum_{i=0}^n r^i = 6558$

We have a formula to compute the sum of the powers of a certain variable, namely

$\displaystyle \sum_{i=0}^n r^i = \cfrac{r^{n+1}-1}{r-1}$

So, the equation becomes

$6\cfrac{r^{n+1}-1}{r-1} = 6558$

The only integer solution to this expression is $n=6, r=3$.

If you want to check the result, we have

$6+6*3+6*3^2+6*3^3+6*3^4+6*3^5+6*3^6 = 6558$

and the last term is

$6*3^6 = 4374$