1.6 - Geometric progressions | Cambridge A-Level Mathematics

Multiplying by the same number repeatedly creates a geometric progression. In this part of 1.6, you will recognise this structure, calculate terms and finite sums, and connect a geometric progression with an arithmetic one. A negative multiplier is allowed, and a common value need not occupy the same position in two sequences.

Recognising a constant ratio

The sequence $12,-6,3,-\frac32,\ldots$ changes sign at every step. Can it still have one consistent rule? Yes: multiply each term by $-\frac12$ . Equal differences are not the only way to build a predictable sequence.

Geometric progression

A geometric progression, or GP, is a sequence in which each term is obtained by multiplying the previous term by a fixed common ratio $r$ .

With first term $a$ , its terms are $a,ar,ar^2,\ldots$ . If consecutive terms are nonzero, divide “next by previous” to find $r$ .

A positive ratio keeps the sign of nonzero terms; a negative ratio alternates it. Ratios between $-1$ and $1$ reduce the magnitude each time. For example, $3,6,12$ is a GP with $r=2$ , whereas $3,6,9$ is an AP and has unequal ratios.

For three consecutive GP terms $a,b,c$ , we have $b=ar$ and $c=ar^2$ , so

b^2=(ar)^2=a(ar^2)=ac.

Conversely, if $a\ne0$ and $b^2=ac$ , choosing $r=b/a$ gives both $b=ar$ and $c=br$ . For nonzero terms this is the usual equal-ratio test.

Worked example 1: an alternating progression

Question. Show that $12,-6,3$ are consecutive terms of a GP, and state its common ratio.

Plan. Divide each term by the preceding one and compare the ratios.

\frac{-6}{12}=-\frac12, \qquad \frac3{-6}=-\frac12.

Both steps use the same multiplier.

Answer. The terms form a GP with $r=-\frac12$ .

Check. The three-term identity gives $(-6)^2=36=12(3)$ . Multiplying $3$ by $-\frac12$ also predicts the next term $-\frac32$ .

Keep the first term, but put a parameter in the other two. Squaring the middle term can give two possible progressions.

Worked example 2: two possible ratios

Question. The numbers $12$ , $k-8$ and $k+1$ , in that order, are consecutive GP terms. Find the possible values of $k$ and the corresponding common ratios.

Plan. Use the three-term identity, solve the resulting quadratic, then check both original ratios.

\begin{aligned} (k-8)^2&=12(k+1)\\ k^2-16k+64&=12k+12\\ k^2-28k+52&=0\\ (k-2)(k-26)&=0. \end{aligned}

For $k=2$ , the terms are $12,-6,3$ , with $r=-\frac12$ .
For $k=26$ , the terms are $12,18,27$ , with $r=\frac32$ .

Answer. $k=2$ , $r=-\frac12$ , or $k=26$ , $r=\frac32$ .

Check. For the second progression, $18/12=27/18=3/2$ ; the first was checked directly above. Both parameter values satisfy the original multiplication rule.

Check zero terms using multiplication rather than dividing by zero. The sequence $5,0,0,\ldots$ is a GP with multiplier $0$ . But $0,0,5$ satisfies $b^2=ac$ and is not a GP: no fixed real multiplier turns the second zero into $5$ . The all-zero sequence does satisfy the multiplication rule, but it does not determine a unique ratio.

Watch out

The equation $b^2=ac$ should be checked against the actual sequence, particularly when zero terms occur. Never calculate a ratio by dividing by zero, and never discard a negative ratio merely because it is negative.

Find and check both possibilities in this new parameter problem.

Terms and finite sums

To reach the fifth term of a GP, multiply the first term by $r$ four times. This gives $ar^4$ . The exponent counts the steps from the first term.

Terms of a GP

For first term $a$ and common ratio $r$ ,

u_n=ar^{n-1}, \qquad n=1,2,3,\ldots.

When $r=0$ , read the sequence directly as $a,0,0,\ldots$ ; its first term remains $a$ .

In the formula booklet.

A term-position problem can sometimes be solved by recognising powers. If $6\times2^{n-1}=192$ , dividing by $6$ gives $2^{n-1}=32=2^5$ , so $n=6$ . Check that the resulting index is a positive integer.

For a sum, write $S_n=a+ar+\cdots+ar^{n-1}$ . Multiplying by $r$ shifts every power along one place. Subtracting cancels all the middle terms:

\begin{aligned} S_n&=a+ar+\cdots+ar^{n-1}\\ rS_n&=\phantom{a+{}}ar+\cdots+ar^{n-1}+ar^n\\ (1-r)S_n&=a-ar^n. \end{aligned}

We can divide by $1-r$ only if $r\ne1$ .

Finite sum of a GP

For a positive integer $n$ ,

S_n=\frac{a(1-r^n)}{1-r}\qquad(r\ne1).

Equivalently, $S_n=a(r^n-1)/(r-1)$ .

In the formula booklet. If $r=1$ , every term is $a$ , so use $S_n=na$ directly.

A finite sum exists for every real ratio. There is no requirement here that the terms shrink.

Worked example 3: positive ratio greater than one

Question. A GP has first term $6$ and common ratio $2$ . Find its fifth term and the sum of its first five terms.

Plan. Use four multiplication steps for the term, and five terms in the sum formula.

\begin{aligned} u_5&=6(2^4)=96\\ S_5&=\frac{6(1-2^5)}{1-2} =\frac{6(-31)}{-1}=186. \end{aligned}

Answer. $u_5=96$ , $S_5=186$ .

Check. Listing the five terms gives $6,12,24,48,96$ , whose sum is $186$ . The fifth term and the whole total are different quantities.

Change only the sign of the ratio. Keep the brackets around it: even powers of $-2$ are positive and odd powers are negative.

Worked example 4: a negative ratio

Question. A GP has first term $6$ and common ratio $-2$ . Find its fifth term and the sum of its first five terms.

Plan. Apply the same formulas with $r=-2$ , retaining the sign inside each power.

\begin{aligned} u_5&=6(-2)^4=96\\ S_5&=\frac{6\{1-(-2)^5\}}{1-(-2)} =\frac{6(33)}3=66. \end{aligned}

Answer. $u_5=96$ , $S_5=66$ .

Check. The terms are $6,-12,24,-48,96$ . Pairing them gives $(6-12)+(24-48)+96=-6-24+96=66$ .

For comparison, first term $6$ and ratio $1$ give $6,6,6,6,6$ , so $u_5=6$ and $S_5=30$ . The quotient formula would contain $0/0$ ; the repeated-term interpretation settles this case immediately.

Watch out

In the term formula use $r^{n-1}$ ; in the finite-sum formula use $r^n$ . Also, $1-(-2)=3$ , not $-1$ . Writing the substitution before simplifying protects both decisions.

Use a different alternating GP.

Linked progressions

Two progressions may share a first term, a later term, or a total. Translate each statement using the rule for the progression it describes. An AP adds a difference; a GP multiplies by a ratio. These parameters have different jobs.

A common term is a value appearing in both sequences. Its positions can differ: the third GP term might equal the sixth AP term. Use different indices, such as $u_n$ and $v_m$ , unless the question explicitly says the positions are the same.

Worked example 5: connecting an AP and a GP

Question. AP A has first term $5$ and common difference $3$ . GP B also has first term $5$ , and its second term is $2$ greater than the second term of A. Find the ratio of B, then determine which term of A equals the third term of B.

Plan. Use the second-term relationship to find the ratio, then equate the GP's known third term to the AP's general term.

The second term of A is $5+3=8$ . Therefore

\begin{aligned} 5r&=8+2=10\\ r&=2\\ v_3&=5(2^2)=20. \end{aligned}

Now find the position of $20$ in A:

5+3(n-1)=20 \quad\Longrightarrow\quad n=6.

Answer. B has ratio $2$ ; its third term equals the sixth term of A, both being $20$ .

Check. A begins $5,8,11,14,17,20$ , while B begins $5,10,20$ . The second terms differ by $2$ , and the common value has the two stated positions.

Keep both progressions, but search for the next larger common value. Since B doubles a positive term, its terms increase: checking them in order will establish which common value comes first.

Worked example 6: the next common value

Question. AP A begins $5,8,11,\ldots$ , and GP B begins $5,10,20,\ldots$ . Find their first common term greater than $20$ , and its position in each progression.

Plan. Generate successive GP candidates and test whether each has a positive integer index in the AP.

The next GP terms after $20$ are $40$ and $80$ . The AP has $u_n=5+3(n-1)=3n+2$ .

\begin{aligned} 3n+2=40&\quad\Longrightarrow\quad n=\frac{38}3 &&\text{not an integer}\\ 3n+2=80&\quad\Longrightarrow\quad n=26 &&\text{valid index}. \end{aligned}

The GP terms are $5,10,20,40,80$ , so $80$ is its fifth term.

Answer. The first common term greater than $20$ is $80$ : term $26$ of A and term $5$ of B.

Check. $5+25(3)=80$ and $5(2^4)=80$ . The only intervening GP term is $40$ , which was rejected, so no smaller candidate was skipped.

The opening alternating sequence is a GP because its multiplier is consistently $-1/2$ . Its changing signs do not affect that conclusion. When comparing progressions, keep the same care over the rule and over which positions the question connects.

Problem-solving

To find the first common value above a bound in an increasing GP and an AP, test the GP candidates in order. A non-integer AP index rejects a candidate. Stop at the first valid one, and state both positions if requested.

Use this reasoning with fresh progressions.

Quick Recap

A GP uses one multiplier $r$ ; nonzero consecutive terms have equal ratios.
For three terms, use $b^2=ac$ , checking any zero cases with the multiplication rule.
$u_n=ar^{n-1}$ gives a term, while $S_n=a(1-r^n)/(1-r)$ gives a finite total when $r\ne1$ . For $r=1$ , use $S_n=na$ .
A common value can occur at different indices in different progressions.

The opening sequence $12,-6,3,-3/2,\ldots$ has the fixed ratio $-1/2$ , so it is geometric despite its alternating signs.

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1.6 - Geometric progressions

Recognising a constant ratio

Terms and finite sums

Linked progressions

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