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Method · Progressions

Working with geometric progressions

A geometric progression multiplies by a fixed common ratio rr at every step. Use Tn=arn1T_n = ar^{n-1} for any term, Sn=a(rn1)r1S_n = \frac{a(r^{n}-1)}{r-1} for a sum, and S=a1rS_\infty = \frac{a}{1-r} when r<1|r|<1.

What this method is for

A geometric progression, or GP, is a sequence in which each term is found by multiplying the term before it by the same fixed number. That fixed number is the common ratio rr.

In Add Math you reach for a GP to model anything that grows or decays by a constant factor rather than by a constant amount, money earning compound interest, a population that doubles each period, the value of a machine that falls to a fixed fraction of its worth each year, or the rebound heights of a bouncing ball that reach the same proportion of the previous drop.

The difference from an arithmetic progression matters: an arithmetic progression adds a fixed amount, so it grows in a straight line, while a geometric progression multiplies by a fixed factor, so it grows or shrinks ever faster. Three formulas cover almost every GP question.

The first finds any single term, the second adds up the first nn terms, and the third gives the total of an unending sum when the terms shrink towards zero.

nth term
Tn=arn1T_n = ar^{\,n-1}
a = first term, r = common ratio
sum of first n terms
Sn=a(rn1)r1,r1S_n = \frac{a\left(r^{n}-1\right)}{r-1}, \quad r \neq 1
sum to infinity
S=a1r,r<1S_\infty = \frac{a}{1-r}, \quad |r| < 1

The sum of the first nn terms also has a mirror-image form, Sn=a(1rn)1rS_n = \frac{a(1-r^{n})}{1-r}, obtained by multiplying the top and bottom of the first version by 1-1. The two are identical in value, but the second keeps every number positive when the common ratio is a proper fraction, so it is the tidier choice for a shrinking GP.

Knowing all three purposes at a glance, one term, a finite sum, an infinite sum, lets you match the formula to the question before you write a single number.

Sum to infinity in action

Take a GP with first term a=8a = 8 and common ratio r=12r = \frac{1}{2}. Because r<1|r| < 1, the terms 8,4,2,1,12,8, 4, 2, 1, \frac{1}{2}, \ldots shrink towards zero, so the running total has a fixed ceiling: S=a1r=8112=812=16S_\infty = \frac{a}{1-r} = \frac{8}{1 - \frac{1}{2}} = \frac{8}{\frac{1}{2}} = 16.

However many terms you add, the total creeps closer to 1616 but never passes it.

When to reach for it

Reach for this method when a list of numbers is multiplied by the same factor at each step rather than having a fixed amount added. Test by dividing neighbouring terms: if T2T1=T3T2\frac{T_2}{T_1} = \frac{T_3}{T_2}, the sequence is geometric and that shared value is rr.

Words such as 'increases by a fixed percentage', 'doubles', 'halves', or 'each bounce reaches a fraction of the last' all point to a GP.

If the question mentions a total that keeps building forever while the terms get smaller, that is the signal for the sum to infinity, which is only valid when r<1|r| < 1. Decide early whether you need a single term, a finite sum, or an infinite sum, because each needs a different formula and mixing them up is the most common way to lose marks on this topic.

A common exam dressing is percentage growth or decay. 'Increases by 20%20\% each year' means multiplying by 1.21.2 every step, so r=1.2r = 1.2; 'loses 20%20\% of its value each year' means multiplying by 0.80.8, so r=0.8r = 0.8.

Translating the percentage into a multiplier is the very first thing to do, because once you hold rr the rest of the question becomes routine substitution.

The common ratio need not be a whole number. It may be a fraction such as 12\frac{1}{2} for a quantity that halves each step, or negative such as 2-2 for terms that swing between positive and negative signs.

A ratio lying strictly between 1-1 and 11 is precisely the case where the sum to infinity exists, so judging the size of rr early also tells you whether a question is even entitled to ask for SS_\infty. When in doubt, write out the first three or four terms; the pattern of multiplication is usually obvious once the numbers sit side by side.

The steps

  1. 1

    Identify a and r

    Write down aa (the first term) and find rr by dividing any term by the one before it: r=T2T1r = \frac{T_2}{T_1}.

  2. 2

    Set up equations if a or r is unknown

    If two terms are given, write each as arn1ar^{\,n-1}, then divide one equation by the other so that aa cancels and only rr is left.

  3. 3

    Solve for r, then a

    Solve the resulting equation for rr, then substitute back into one term equation to find aa.

  4. 4

    Choose the right formula

    Use TnT_n for a single term, SnS_n for the sum of nn terms, and SS_\infty only when r<1|r| < 1.

  5. 5

    Substitute and simplify

    Put the numbers in carefully, working the powers out one line at a time, and state the final value.

  6. 6

    Check

    List a few terms with your aa and rr, or substitute back into a given term, to confirm the answer is sensible.

Worked example

Q1[5 marks]

In a geometric progression, the 2nd term is 6 and the 5th term is 48. Find the first term aa, the common ratio rr, and the sum of the first 6 terms.

Show worked solution

Write each given term using Tn=arn1T_n = ar^{\,n-1}.

T2=ar=6T_2 = ar = 6
T5=ar4=48T_5 = ar^{4} = 48

Divide the second equation by the first so that aa cancels, this is the key move that isolates the ratio:

ar4ar=486    r3=8    r=2\frac{ar^{4}}{ar} = \frac{48}{6} \;\Rightarrow\; r^{3} = 8 \;\Rightarrow\; r = 2

Because r3=8r^{3} = 8 has the single real cube root r=2r = 2, there is no ambiguity here. Substitute r=2r = 2 back into ar=6ar = 6:

a(2)=6    a=3a(2) = 6 \;\Rightarrow\; a = 3

Since r=2>1r = 2 > 1, the terms are growing, so the form Sn=a(rn1)r1S_n = \frac{a(r^{n}-1)}{r-1} keeps the arithmetic positive. Take n=6n = 6:

S6=3(261)21=3(641)1=3×63=189S_6 = \frac{3\left(2^{6}-1\right)}{2-1} = \frac{3(64-1)}{1} = 3 \times 63 = 189

So a=3a = 3, r=2r = 2, and the sum of the first 6 terms is 189189. As a check, the six terms are 3,6,12,24,48,963, 6, 12, 24, 48, 96; adding them gives 189189, and the 2nd and 5th terms are indeed 66 and 4848, exactly as the question stated.

Common pitfalls

  • Using rnr^{n} instead of rn1r^{\,n-1} in the term formula, the first term needs zero multiplications by rr, so the power is (n1)(n-1).
  • Adding instead of dividing when finding rr; the common ratio comes from T2T1\frac{T_2}{T_1}, not T2T1T_2 - T_1.
  • Applying the sum-to-infinity formula when r1|r| \geq 1, where the sum does not converge to a finite value.
  • Losing a negative common ratio: if the terms alternate in sign, rr is negative and its powers change sign accordingly.
  • Mixing up SnS_n and TnT_n: a 'sum' question wants a total, a 'term' question wants a single value.

How a teacher helps

Most GP mistakes trace back to one moment, a student subtracts to find rr as though it were an arithmetic progression, or reaches for the sum to infinity when r|r| is not less than one. In one-to-one lessons our teachers watch exactly where you set up the ratio and the power, and correct the slip the instant it appears, so the wrong instinct never takes hold.

We also make you write the 'divide one equation by the other' line clearly, because that is where the common ratio is cleanly found and where method marks are earned. When a ratio turns out to be a fraction or a negative number, we slow down and rehearse the signs and powers until they feel natural.

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Frequently asked questions

How do I tell a geometric progression from an arithmetic one?

Divide neighbouring terms. If T2T1=T3T2\frac{T_2}{T_1} = \frac{T_3}{T_2}, the sequence is geometric with common ratio rr.

If instead the difference between terms is constant, it is arithmetic. A GP multiplies; an AP adds.

When can I use the sum to infinity?

Only when the common ratio satisfies r<1|r| < 1, so the terms shrink towards zero. Then S=a1rS_\infty = \frac{a}{1-r}.

If r1|r| \geq 1 the terms do not shrink and the infinite sum has no finite value.

How do I find the common ratio from two non-adjacent terms?

Write each term as arn1ar^{\,n-1} and divide one equation by the other; aa cancels and you are left with a power of rr. For example ar4ar=r3\frac{ar^{4}}{ar} = r^{3}, so a known value of r3r^{3} gives rr directly.

Can the common ratio be a fraction or negative?

Yes. A fraction such as r=12r = \frac{1}{2} describes a quantity that halves each step, and a negative ratio such as r=2r = -2 makes the terms alternate in sign.

Both use the same formulas; just carry the sign and the fraction carefully through each power.

Which geometric progression formulas must I memorise?

Memorise Tn=arn1T_n = ar^{\,n-1} for the nnth term, Sn=a(rn1)r1S_n = \frac{a(r^{n}-1)}{r-1} for the sum of the first nn terms, and S=a1rS_\infty = \frac{a}{1-r} for the sum to infinity. Because marking is analytic, showing the formula and substitution earns method marks even if the final value slips.

Source:SRC-DSKP-EN

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

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