iGCSE (2021 Edition)

# 9.05 Geometric series

Lesson

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,

### Summing $n$n terms in a geometric sequence

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:

$S_n=u_1+u_1r+u_1r^2+u_1r^3+...+u_1r^{n-1}$Sn=u1+u1r+u1r2+u1r3+...+u1rn1

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:

$rS_n=u_1r+u_1r^2+u_1r^3+...+u_1r^{n-1}+u_1r^n$rSn=u1r+u1r2+u1r3+...+u1rn1+u1rn

Then, by carefully subtracting $rS_n$rSn from $S_n$Sn term by term, we see that all of the middle terms disappear:

$S_n-rS_n=u_1+\left(u_1r-u_1r\right)+\left(u_1r^2-u_1r^2\right)+...+\left(u_1r^{n-1}-u_1r^{n-1}\right)-u_1r^n$SnrSn=u1+(u1ru1r)+(u1r2u1r2)+...+(u1rn1u1rn1)u1rn

This means that:

$S_n-rS_n=u_1-u_1r^n$SnrSn=u1u1rn

and when common factors are taken out on both sides of this equation, we find:

$S_n\left(1-r\right)=u_1\left(1-r^n\right)$Sn(1r)=u1(1rn)

Finally, by dividing both sides by $\left(1-r\right)$(1r) (excluding the trivial case of $r=1$r=1) we reveal the geometric sum formula:

$S_n=\frac{u_1\left(1-r^n\right)}{1-r}$Sn=u1(1rn)1r

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:

$S_n=\frac{u_1\left(r^n-1\right)}{r-1}$Sn=u1(rn1)r1

Geometric series

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:

$S_n=\frac{u_1\left(1-r^n\right)}{1-r}$Sn=u1(1rn)1r, for $r<1$r<1 or $S_n=\frac{u_1\left(r^n-1\right)}{r-1}$Sn=u1(rn1)r1, 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​

#### Worked example

##### Example 1

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 $S_n=\frac{u_1\left(r^n-1\right)}{r-1}$Sn=u1(rn1)r1 and rearrange.

Do: $u_1=5$u1=5, $r=2$r=2 and $S_n=5115$Sn=5115, so we have:

 $S_n$Sn​ $=$= $\frac{u_1\left(r^n-1\right)}{r-1}$u1​(rn−1)r−1​ $5115$5115 $=$= $\frac{5\left(2^n-1\right)}{2-1}$5(2n−1)2−1​ Substitute values into formula Hence,$5\left(2^n-1\right)$5(2n−1) $=$= $5115$5115 Simplify fraction and bring unknown to left-hand side $2^n-1$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.

#### Practice questions

##### Question 1

Consider the series $4+12+36$4+12+36 ...

Find the sum of the first $12$12 terms.

##### Question 2

Consider the series $5+\frac{5}{2}+\frac{5}{4}$5+52+54 ...

1. Find the common ratio, $r$r.

2. Find the sum of the first $9$9 terms, rounding your answer to 1 decimal place.

##### Question 3

Consider the series $4-12+36-\text{. . . }-708588$412+36. . . 708588.

1. Solve for $n$n, the number of terms in the series.

2. Find the sum of the series.

##### Question 4

Average annual salaries are expected to increase by $4$4 percent each year. If the average annual salary this year is found to be $\$4000040000:

1. Calculate the expected average annual salary in $6$6 years, correct to the nearest cent.

2. 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.

### Outcomes

#### 0606C12.3B

Recognise geometric progressions.

#### 0606C12.4B

Use the formulae for the nth term and for the sum of the first n terms to solve problems involving geometric progressions.