If the first term of a geometric sequence is $u_1$u1 and the common ratio is $r$r, then the sequence is given by:
$u_1,u_1r,u_1r^2,u_1r^3,u_1r^4,\ldots$u1,u1r,u1r2,u1r3,u1r4,…
Suppose we wish to add the first $n$n terms of this sequence. This will form a geometric series. We could write the sum as:
$S_n=u_1+u_1r+u_1r^2+u_1r^3+...+u_1r^{n1}$Sn=u1+u1r+u1r2+u1r3+...+u1rn−1
We saw a nifty trick for finding the formula for an arithmetic series by adding two sums together and pairing up terms. We will use a similar method here. If we multiply both sides of our geometric sum by the common ratio $r$r we see that:
$rS_n=u_1r+u_1r^2+u_1r^3+...+u_1r^{n1}+u_1r^n$rSn=u1r+u1r2+u1r3+...+u1rn−1+u1rn
Then, by carefully subtracting $rS_n$rSn from $S_n$Sn term by term, we see that all of the middle terms disappear:
$S_nrS_n=u_1+\left(u_1ru_1r\right)+\left(u_1r^2u_1r^2\right)+...+\left(u_1r^{n1}u_1r^{n1}\right)u_1r^n$Sn−rSn=u1+(u1r−u1r)+(u1r2−u1r2)+...+(u1rn−1−u1rn−1)−u1rn
This means that:
$S_nrS_n=u_1u_1r^n$Sn−rSn=u1−u1rn
and when common factors are taken out on both sides of this equation, we find:
$S_n\left(1r\right)=u_1\left(1r^n\right)$Sn(1−r)=u1(1−rn)
Finally, by dividing both sides by $\left(1r\right)$(1−r) (excluding the trivial case of $r=1$r=1) we reveal the geometric sum formula:
$S_n=\frac{u_1\left(1r^n\right)}{1r}$Sn=u1(1−rn)1−r
An extra step, multiplying the numerator and denominator by $1$−1, reveals a different form for $S_n$Sn. Both formulas will work in any situation, particularly when using a calculator. This form is generally easier to manage when the common ratio is greater than $r=1$r=1:
$S_n=\frac{u_1\left(r^n1\right)}{r1}$Sn=u1(rn−1)r−1
For any geometric sequence with starting value $u_1$u1 and common ratio $r$r, we can find the sum of the first $n$n terms, using:
$S_n=\frac{u_1\left(1r^n\right)}{1r}$Sn=u1(1−rn)1−r, for $r<1$r<1 or $S_n=\frac{u_1\left(r^n1\right)}{r1}$Sn=u1(rn−1)r−1, convenient if $r>1$r>1
The formulas for $S_n$Sn exclude the case for $r=1$r=1. For the case where $r=1$r=1, then the sequence becomes $u_1,u_1,u_1,u_1,\ldots$u1,u1,u1,u1,….
Is this a geometric sequence with $r=1$r=1 or an arithmetic sequence with $d=0$d=0? Either way, every term is clearly identical. Hence, the sum of the first $n$n terms is:
$S_n$Sn  $=$=  $u_1+u_1+u_1+...+u_1$u1+u1+u1+...+u1  ($n$n times) 
$S_n$Sn  $=$=  $nu_1$nu1 
If the sum for the first $n$n terms of the geometric sequence $5,10,20,\ldots$5,10,20,… is $5115$5115, find $n$n.
Think: We have an increasing geometric sequence. State $u_1$u1, $r$r and $S_n$Sn, then substitute into the formula $S_n=\frac{u_1\left(r^n1\right)}{r1}$Sn=u1(rn−1)r−1 and rearrange.
Do: $u_1=5$u1=5, $r=2$r=2 and $S_n=5115$Sn=5115, so we have:
$S_n$Sn  $=$=  $\frac{u_1\left(r^n1\right)}{r1}$u1(rn−1)r−1  
$5115$5115  $=$=  $\frac{5\left(2^n1\right)}{21}$5(2n−1)2−1 
Substitute values into formula

Hence,$5\left(2^n1\right)$5(2n−1)  $=$=  $5115$5115 
Simplify fraction and bring unknown to lefthand side

$2^n1$2n−1  $=$=  $1023$1023 
Divide both sides by $5$5

$2^n$2n  $=$=  $1024$1024 
Add $1$1 to both sides

$\therefore n$∴n  $=$=  $10$10 
Solve for $n$n, using guess and check, technology or logarithms.

Consider the series $4+12+36$4+12+36 ...
Find the sum of the first $12$12 terms.
Consider the series $5+\frac{5}{2}+\frac{5}{4}$5+52+54 ...
Find the common ratio, $r$r.
Find the sum of the first $9$9 terms, rounding your answer to 1 decimal place.
Consider the series $412+36\text{. . . }708588$4−12+36−. . . −708588.
Solve for $n$n, the number of terms in the series.
Find the sum of the series.
Average annual salaries are expected to increase by $4$4 percent each year. If the average annual salary this year is found to be $\$40000$$40000:
Calculate the expected average annual salary in $6$6 years, correct to the nearest cent.
This year, Vincent starts at a new job in which he will receive the average annual salary for each year of his employment. Over the coming $6$6 years (including this year) he plans to save half of each year’s annual salary.
What will be his total savings over these $6$6 years? Give your answer correct to the nearest cent.
Recognise geometric progressions.
Use the formulae for the nth term and for the sum of the first n terms to solve problems involving geometric progressions.