Mathematics 9709 · Pure Mathematics 1

Geometric progressions

A geometric progression (GP) multiplies by the same number each step, like $2,\,6,\,18,\,54,\dots$ (times $3$ ).
That fixed multiplier is the common ratio $r$ . The start is the first term $a$ .
Everything in the topic comes from those two numbers.

Given two terms? Divide the two equations. The $a$ cancels, and you are left with just a power of $r$ . Then take the root.

AP vs GP: gap or multiplier

First decide which kind it is, because they use different formulas, by looking at how you get from one term to the next:

	Arithmetic (AP)	Geometric (GP)
Step rule	ADD the same amount	MULTIPLY by the same amount
Constant	common difference $d$	common ratio $r$
Example	$5,\,8,\,11,\,14$	$2,\,6,\,18,\,54$
Check it by	subtracting neighbours	dividing neighbours

To test a GP, divide any term by the one before it. If the answer is always the same, that answer is the ratio $r$ .

Three consecutive terms: the middle-term shortcut

When a question gives three terms in a row $a,\,b,\,c$ , the middle one $b$ links the other two. AP and GP link them in different ways:

In an AP the gaps are equal, $b-a=c-b$ . So the middle is the average: $2b=a+c$ .
In a GP the ratios are equal, $\tfrac{b}{a}=\tfrac{c}{b}$ . So the middle is the geometric mean: $b^2=ac$ .

How to remember: add for AP (“middle is the average”), multiply for GP (“middle squared is the product”). So if $x,\,4,\,x+6$ are three GP terms in a row, $4^2=x(x+6)$ gives $x^2+6x-16=0$ , so $x=2$ or $x=-8$ .

A GP multiplies by the same number each step (here ×3), so the jumps get bigger and bigger, unlike an AP's equal gaps.

The term and the sum

To get to the $n$ th term, you multiply by $r$ a total of $(n-1)$ times.
It is the same off-by-one as an AP. You just multiply instead of add:

$u_n=ar^{\,n-1}$

For the sum of the first $n$ terms there is one formula. It can be written two ways that mean the same:

$S_n=\frac{a(1-r^{\,n})}{1-r}=\frac{a(r^{\,n}-1)}{r-1}$

The two forms are equal. Pick the one that keeps the numbers positive. Use the left form when $r<1$ , and the right form when $r>1$ .
Both are in your MF19 formula list.
Neither works when $r=1$ . But then the list is just $a,a,a,\dots$ , and the sum is $na$ .

Where the sum formula comes from (multiply and subtract)

Write out the sum. Then write $r$ times the sum right underneath, moved along by one term:

$S_n=a+ar+ar^2+\dots+ar^{\,n-1}$

$rS_n=\ \ \ \ ar+ar^2+\dots+ar^{\,n-1}+ar^{\,n}$

Subtract the second line from the first. Every middle term cancels. Only the very first term and the very last term are left:

$S_n-rS_n=a-ar^{\,n}$

Take out a common factor on each side, then divide by $(1-r)$ :

$S_n(1-r)=a(1-r^{\,n}) \;\Rightarrow\; S_n=\frac{a(1-r^{\,n})}{1-r}$

Multiply the top and the bottom by $-1$ . This gives the other form, $\dfrac{a(r^{\,n}-1)}{r-1}$ . That is why you can use either one.

Worked examples

Two terms $b$ steps apart? Divide their equations so $a$ cancels. You are left with just a power of $r$ . An odd root (a cube root) gives one value. An even root (a square root) gives $\pm$ , so check the question to see which sign to keep.

Worked example

Two terms given: find $r$ and $a$

The 2nd term of a GP is $12$ and the 5th is $40.5$ .

Write each with the term formula: $ar=12$ and $ar^4=40.5$ .
Divide the second by the first so $a$ cancels: $\dfrac{ar^4}{ar}=\dfrac{40.5}{12}\Rightarrow r^3=\tfrac{27}{8}$ .
Cube-root (one value): $r=\sqrt[3]{\tfrac{27}{8}}=\tfrac32$ , then back-substitute into $ar=12$ : $a=\dfrac{12}{3/2}=8$ .

Answer

a=8,\quad r=\tfrac32

Worked example

Find the sum of the first 12 terms of $3 + 6 + 12 + 24 + \dots$

From the start and the multiplier, $a=3$ and $r=2$ . Since $r>1$ , use the right-hand form to keep the numbers positive:

$S_{12}=\frac{a(r^{12}-1)}{r-1}=\frac{3(2^{12}-1)}{2-1}=3(4096-1)$

Answer

S_{12}=12285

Worked example

Real exam: three terms in a row give the ratio

The fifth, sixth and seventh terms of a GP are $8k$ , $-12$ and $2k$ respectively, with $k$ negative. Find the sum to infinity.

Step 1. For three GP terms in a row, the middle one squared equals the other two multiplied ( $b^2=ac$ ):

$(-12)^2=(8k)(2k)\Rightarrow 144=16k^2\Rightarrow k^2=9$

Step 2. So $k=\pm3$ . The question says $k$ is negative, so $k=-3$ . Now the three terms are real numbers: $8k=-24$ , then $-12$ , then $2k=-6$ . The ratio is a term divided by the one before it:

$r=\frac{-12}{-24}=\tfrac12$

Step 3. Find the first term by going back from the 5th term, $ar^4=-24$ , so $a=\dfrac{-24}{(1/2)^4}=-24\times16=-384$ . Since $|r|<1$ , use the sum to infinity:

$S_\infty=\frac{a}{1-r}=\frac{-384}{1-\tfrac12}=\frac{-384}{\tfrac12}$

Answer

S_\infty=-768

(9709/11 Jun 2021 Q5) A negative $a$ is fine. The formula works with any sign, as long as $|r|<1$ .

Where the marks go

Common mistake

The

n

th term is

ar^{\,n-1}

, not

ar^{\,n}

. The power is one less than the term number, so the 5th term is

ar^4

. This off-by-one is the most common GP slip.

Common mistake

An even root can give two values of

r

(for example

r^2=4

gives

r=\pm2

). Read the question. A clue like “all terms positive” or

r>\tfrac12

tells you which root to keep. (9709/12 Nov 2021 Q6)

Now you try

The 2nd term of a GP is $54$ and the sum to infinity is $243$ . The common ratio is greater than $\tfrac12$ . Find the tenth term, in exact form. (9709/12 Nov 2021 Q6)

(The sum to infinity is the next topic, Infinite geometric series. When $|r|<1$ it equals $\dfrac{a}{1-r}$ .)

Your turn— tap to reveal the worked answer (9709/12 Nov 2021 Q6)

Two facts: $ar=54$ and $\dfrac{a}{1-r}=243$ . From the first, $a=\dfrac{54}{r}$ . Put that into the second:

$\frac{54/r}{1-r}=243\;\Rightarrow\;54=243\,r(1-r)$

Multiply out and tidy into a quadratic, then divide by 27:

$243r^2-243r+54=0\;\Rightarrow\;9r^2-9r+2=0$

Solve the quadratic. The calculator gives the roots. To write the factors, see the Quadratics chapter:

$(3r-1)(3r-2)=0\;\Rightarrow\;r=\tfrac13\ \text{or}\ \tfrac23$

The rule $r>\tfrac12$ rejects $\tfrac13$ . So $r=\tfrac23$ and $a=\dfrac{54}{2/3}=81$ . Then:

$u_{10}=ar^9=81\left(\tfrac23\right)^{9}$

Answer

u_{10}=\dfrac{512}{243}

PreviousArithmetic progressions NextInfinite geometric series

Geometric progressions

A geometric progression (GP) multiplies by the same number each step, like $2,\,6,\,18,\,54,\dots$ (times $3$ ).
That fixed multiplier is the common ratio $r$ . The start is the first term $a$ .
Everything in the topic comes from those two numbers.

Given two terms? Divide the two equations. The $a$ cancels, and you are left with just a power of $r$ . Then take the root.

AP vs GP: gap or multiplier

First decide which kind it is, because they use different formulas, by looking at how you get from one term to the next:

	Arithmetic (AP)	Geometric (GP)
Step rule	ADD the same amount	MULTIPLY by the same amount
Constant	common difference $d$	common ratio $r$
Example	$5,\,8,\,11,\,14$	$2,\,6,\,18,\,54$
Check it by	subtracting neighbours	dividing neighbours

To test a GP, divide any term by the one before it. If the answer is always the same, that answer is the ratio $r$ .

Three consecutive terms: the middle-term shortcut

When a question gives three terms in a row $a,\,b,\,c$ , the middle one $b$ links the other two. AP and GP link them in different ways:

In an AP the gaps are equal, $b-a=c-b$ . So the middle is the average: $2b=a+c$ .
In a GP the ratios are equal, $\tfrac{b}{a}=\tfrac{c}{b}$ . So the middle is the geometric mean: $b^2=ac$ .

A GP multiplies by the same number each step (here ×3), so the jumps get bigger and bigger, unlike an AP's equal gaps.

The term and the sum

To get to the $n$ th term, you multiply by $r$ a total of $(n-1)$ times.
It is the same off-by-one as an AP. You just multiply instead of add:

$u_n=ar^{\,n-1}$

For the sum of the first $n$ terms there is one formula. It can be written two ways that mean the same:

$S_n=\frac{a(1-r^{\,n})}{1-r}=\frac{a(r^{\,n}-1)}{r-1}$

The two forms are equal. Pick the one that keeps the numbers positive. Use the left form when $r<1$ , and the right form when $r>1$ .
Both are in your MF19 formula list.
Neither works when $r=1$ . But then the list is just $a,a,a,\dots$ , and the sum is $na$ .

Where the sum formula comes from (multiply and subtract)

Write out the sum. Then write $r$ times the sum right underneath, moved along by one term:

$S_n=a+ar+ar^2+\dots+ar^{\,n-1}$

$rS_n=\ \ \ \ ar+ar^2+\dots+ar^{\,n-1}+ar^{\,n}$

Subtract the second line from the first. Every middle term cancels. Only the very first term and the very last term are left:

$S_n-rS_n=a-ar^{\,n}$

Take out a common factor on each side, then divide by $(1-r)$ :

$S_n(1-r)=a(1-r^{\,n}) \;\Rightarrow\; S_n=\frac{a(1-r^{\,n})}{1-r}$

Multiply the top and the bottom by $-1$ . This gives the other form, $\dfrac{a(r^{\,n}-1)}{r-1}$ . That is why you can use either one.

Worked examples

Worked example

Two terms given: find $r$ and $a$

The 2nd term of a GP is $12$ and the 5th is $40.5$ .

Write each with the term formula: $ar=12$ and $ar^4=40.5$ .
Divide the second by the first so $a$ cancels: $\dfrac{ar^4}{ar}=\dfrac{40.5}{12}\Rightarrow r^3=\tfrac{27}{8}$ .
Cube-root (one value): $r=\sqrt[3]{\tfrac{27}{8}}=\tfrac32$ , then back-substitute into $ar=12$ : $a=\dfrac{12}{3/2}=8$ .

Answer

a=8,\quad r=\tfrac32

Worked example

Find the sum of the first 12 terms of $3 + 6 + 12 + 24 + \dots$

From the start and the multiplier, $a=3$ and $r=2$ . Since $r>1$ , use the right-hand form to keep the numbers positive:

$S_{12}=\frac{a(r^{12}-1)}{r-1}=\frac{3(2^{12}-1)}{2-1}=3(4096-1)$

Answer

S_{12}=12285

Worked example

Real exam: three terms in a row give the ratio

The fifth, sixth and seventh terms of a GP are $8k$ , $-12$ and $2k$ respectively, with $k$ negative. Find the sum to infinity.

Step 1. For three GP terms in a row, the middle one squared equals the other two multiplied ( $b^2=ac$ ):

$(-12)^2=(8k)(2k)\Rightarrow 144=16k^2\Rightarrow k^2=9$

Step 2. So $k=\pm3$ . The question says $k$ is negative, so $k=-3$ . Now the three terms are real numbers: $8k=-24$ , then $-12$ , then $2k=-6$ . The ratio is a term divided by the one before it:

$r=\frac{-12}{-24}=\tfrac12$

Step 3. Find the first term by going back from the 5th term, $ar^4=-24$ , so $a=\dfrac{-24}{(1/2)^4}=-24\times16=-384$ . Since $|r|<1$ , use the sum to infinity:

$S_\infty=\frac{a}{1-r}=\frac{-384}{1-\tfrac12}=\frac{-384}{\tfrac12}$

Answer

S_\infty=-768

(9709/11 Jun 2021 Q5) A negative $a$ is fine. The formula works with any sign, as long as $|r|<1$ .

Where the marks go

Common mistake

The

n

th term is

ar^{\,n-1}

, not

ar^{\,n}

. The power is one less than the term number, so the 5th term is

ar^4

. This off-by-one is the most common GP slip.

Common mistake

An even root can give two values of

r

(for example

r^2=4

gives

r=\pm2

). Read the question. A clue like “all terms positive” or

r>\tfrac12

tells you which root to keep. (9709/12 Nov 2021 Q6)

Now you try

The 2nd term of a GP is $54$ and the sum to infinity is $243$ . The common ratio is greater than $\tfrac12$ . Find the tenth term, in exact form. (9709/12 Nov 2021 Q6)

(The sum to infinity is the next topic, Infinite geometric series. When $|r|<1$ it equals $\dfrac{a}{1-r}$ .)

Your turn— tap to reveal the worked answer (9709/12 Nov 2021 Q6)

Two facts: $ar=54$ and $\dfrac{a}{1-r}=243$ . From the first, $a=\dfrac{54}{r}$ . Put that into the second:

$\frac{54/r}{1-r}=243\;\Rightarrow\;54=243\,r(1-r)$

Multiply out and tidy into a quadratic, then divide by 27:

$243r^2-243r+54=0\;\Rightarrow\;9r^2-9r+2=0$

Solve the quadratic. The calculator gives the roots. To write the factors, see the Quadratics chapter:

$(3r-1)(3r-2)=0\;\Rightarrow\;r=\tfrac13\ \text{or}\ \tfrac23$

The rule $r>\tfrac12$ rejects $\tfrac13$ . So $r=\tfrac23$ and $a=\dfrac{54}{2/3}=81$ . Then:

$u_{10}=ar^9=81\left(\tfrac23\right)^{9}$

Answer

u_{10}=\dfrac{512}{243}