Nth Term of a GP — Formula and Worked Examples

#Algebra
TL;DR
The nth term of a GP (geometric progression) is found with an=arn−1a_n = ar^{n-1}, where aa is the first term and rr is the common ratio. This article gives you the formula, a clean derivation, the term-from-the-end version, six worked examples, and the slips that cost marks — so you can find any term without writing out the whole sequence
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Madhukar VLast updated on July 1, 20268 min read

What is the nth Term of a GP?

The nth term of a GP is the value of the term in position nn of a geometric progression, given by the formula an=arn−1a_n = ar^{n-1}. It lets you jump straight to any term without listing every term before it.

First, the terms in that formula. A geometric progression is a sequence where each term is the previous one multiplied by a fixed number, the common ratio rr. In 3,6,12,24,…3, 6, 12, 24, \dots each term is double the last, so r=2r = 2 and the first term a=3a = 3. You find rr by dividing any term by the one before it:

r=a2a1=63=2r = \frac{a_2}{a_1} = \frac{6}{3} = 2

A general GP is written:

a,;ar,;ar2,;ar3,;…a, ; ar, ; ar^2, ; ar^3, ; \dots

This is the multiplicative cousin of the arithmetic progressions family, where you add a fixed amount instead of multiplying. It is also covered, under its other name, in geometric sequence.

The nth-term formula and where it comes from

The formula is:

an=arn−1a_n = ar^{n-1}

where aa is the first term, rr is the common ratio, nn is the position of the term you want, and ana_n is its value.

The derivation is just counting how many times you multiply by rr. To reach the first term you multiply by rr zero times; the second term, once; the nnth term, (n−1)(n-1) times:

a1=aa_1 = a

a2=a⋅r=ar1a_2 = a \cdot r = ar^1

a3=ar⋅r=ar2a_3 = ar \cdot r = ar^2

an=arn−1a_n = ar^{n-1}

The exponent is (n−1)(n-1), not nn, because the first term already exists before any multiplication happens. That single fact is where most errors live.

The nth term from the end

Sometimes you want a term counted from the last term of a finite GP, not the first. If the GP has nn terms and last term ll, the kkth term from the end is:

an−k+1=l⋅(1r)k−1a_{n-k+1} = l \cdot \left(\frac{1}{r}\right)^{k-1}

You divide by rr as you walk backwards, which is the same as multiplying by 1r\dfrac{1}{r} — note the stacked fraction, not "1 over r" written inline.

Once you can find any term, the natural next questions are the sums — handled in sum of a GP for a finite GP and sum of an infinite GP when the terms shrink toward zero.

Examples of the Nth Term of a GP

Six problems, easier to harder, all using an=arn−1a_n = ar^{n-1}.

Example 1

Find the common ratio of 5,15,45,135,…5, 15, 45, 135, \dots

Divide a term by the one before it: 155=3\dfrac{15}{5} = 3 and 4515=3\dfrac{45}{15} = 3.

Final answer: r=3r = 3.

Example 2

Find the 6th term of 2,6,18,54,…2, 6, 18, 54, \dots

A frequent first move is to write a6=arn=2⋅36a_6 = ar^n = 2 \cdot 3^6.

Why that is wrong. Using rnr^n instead of rn−1r^{n-1} multiplies by the ratio one extra time, landing you on the 7th term, not the 6th. To reach the 6th term from the first, you multiply by rr only five times.

The correct method. Here a=2a = 2, r=3r = 3, n=6n = 6:

a6=arn−1=2⋅36−1=2⋅35a_6 = ar^{n-1} = 2 \cdot 3^{6-1} = 2 \cdot 3^5

a6=2⋅243=486a_6 = 2 \cdot 243 = 486

Final answer: 486486. The exponent is n−1n-1, every time.

Example 3

Find the 8th term of 1,2,4,8,…1, 2, 4, 8, \dots

Here a=1a = 1, r=2r = 2, n=8n = 8:

a8=1⋅28−1=27=128a_8 = 1 \cdot 2^{8-1} = 2^7 = 128

Final answer: 128128.

Example 4

Find the 5th term of a GP with first term 22 and common ratio 13\dfrac{1}{3}.

Here a=2a = 2, r=13r = \dfrac{1}{3}, n=5n = 5:

a5=2⋅(13)5−1=2⋅(13)4a_5 = 2 \cdot \left(\frac{1}{3}\right)^{5-1} = 2 \cdot \left(\frac{1}{3}\right)^{4}

a5=2⋅181=281a_5 = 2 \cdot \frac{1}{81} = \frac{2}{81}

Final answer: 281\dfrac{2}{81}. When ∣r∣<1|r| < 1 the terms shrink, which is what powers the infinite-sum case.

Example 5

In the GP 3,6,12,…3, 6, 12, \dots, which term equals 192192?

Here a=3a = 3, r=2r = 2, an=192a_n = 192. Solve for nn:

192=3⋅2n−1192 = 3 \cdot 2^{n-1}

64=2n−164 = 2^{n-1}

26=2n−12^6 = 2^{n-1}

n−1=6n - 1 = 6, so n=7n = 7.

Final answer: the 7th term.

Example 6

A bacterial culture starts with 500500 cells and triples every hour. How many cells after 44 hours?

The hourly counts form a GP: a=500a = 500, r=3r = 3. After 44 hours is the 5th term (n=5n = 5, since the start counts as term 1):

a5=500⋅35−1=500⋅34a_5 = 500 \cdot 3^{5-1} = 500 \cdot 3^4

a5=500⋅81=40500a_5 = 500 \cdot 81 = 40500

Final answer: 40,50040{,}500 cells. Watch the indexing — "after 4 hours" is the 5th term when hour zero is the first.

Where the nth term of a GP earns its keep

A GP captures anything that grows or shrinks by a constant factor rather than a constant amount, and the nth-term formula is how you predict where it lands.

  • Compound interest: Money growing at a fixed percentage each period follows a GP; the nth term gives the balance after nn periods without year-by-year addition.

  • Population and decay: Bacterial growth, radioactive decay, and drug clearance all multiply by a fixed factor each step — the nth term is the prediction tool.

  • Computing and data: Storage doubling, signal halving, and the depth of binary structures are all geometric — which is why the chessboard legend feels relevant to anyone who works with exponential growth.

The systematic study of such ratios runs back to Euclid, whose Elements (around 300 BCE) treats geometric progressions formally. The reason the formula matters is the rice-grain reason: geometric growth outruns intuition fast, and arn−1ar^{n-1} is the only way to see exactly how fast.

Common Mistakes With the Nth Term of a GP

Mistake 1: Using rnr^n instead of rn−1r^{n-1}

Where it slips in: Computing a specific term. The rusher reads "6th term" and raises rr to the 6th power.

Don't do this: a6=ar6a_6 = ar^6.

The correct way: a6=ar5a_6 = ar^5. You multiply by the common ratio one fewer time than the term number, because the first term is already there before any multiplication. The habit that fixes it: write the exponent as (n−1)(n-1) in symbols before you put a number in.

Mistake 2: Finding the common ratio by subtracting

Where it slips in: A student fresh off arithmetic progressions reaches for subtraction, the way you find a common difference. The memorizer who learned AP first is most at risk.

Don't do this: For 4,12,36,…4, 12, 36, \dots write r=12−4=8r = 12 - 4 = 8.

The correct way: r=124=3r = \dfrac{12}{4} = 3. The common ratio comes from division, not subtraction. If dividing consecutive terms gives the same number each time, that number is rr; if it does not, the sequence is not a GP at all.

Mistake 3: Mishandling a fractional or negative ratio

Where it slips in: When rr is a fraction or negative, the second-guesser drops the bracket and powers only the numerator, or loses track of the alternating sign.

Don't do this: Write (13)4\left(\frac{1}{3}\right)^4 as 13=13\frac{1}{3^{}} = \frac{1}{3}, powering only the bottom partially, or write (−2)3(-2)^3 as 232^3.

The correct way: Power the whole ratio: (13)4=181\left(\frac{1}{3}\right)^4 = \dfrac{1}{81}, and keep the sign inside the bracket: (−2)3=−8(-2)^3 = -8. A negative rr makes the terms alternate in sign — that is correct behaviour, not an error to "fix." This is the same discipline-with-inputs that, scaled up, mattered when NASA lost the Mars Climate Orbiter to a single mishandled quantity.

Conclusion

  • The nth term of a GP is an=arn−1a_n = ar^{n-1} — first term times the ratio raised to one less than the position.

  • The exponent is (n−1)(n-1), not nn, because the first term exists before any multiplication.

  • The common ratio rr is found by dividing consecutive terms, never subtracting.

  • A negative ratio alternates signs; a fractional ratio shrinks the terms — both are normal.

  • The formula is the practical tool behind compound interest, population growth, and decay.

Quick Self-Check — Try These

Work through each, then revisit the linked section if you stall.

  1. Find the 7th term of 3,6,12,24,…3, 6, 12, 24, \dots (Revisit Example 2 and Mistake 1.)

  2. Find the common ratio of 81,27,9,3,…81, 27, 9, 3, \dots (Revisit Example 1 and Mistake 2.)

  3. Which term of 2,8,32,…2, 8, 32, \dots equals 512512? (Revisit Example 5.)

To work through more GP problems with a teacher, explore Bhanzu's algebra tutor or browse math classes online. Want a live Bhanzu trainer to walk through more nth-term problems? Book a free demo class.

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Frequently Asked Questions

What is the formula for the nth term of a GP?
an=arn−1a_n = ar^{n-1}, where aa is the first term, rr is the common ratio, and nn is the term's position.
How is the nth term of a GP different from the nth term of an AP?
The GP formula uses a power of the common ratio, arn−1ar^{n-1}, because each step multiplies. The AP formula uses a multiple of the common difference, a+(n−1)da + (n-1)d, because each step adds. The arithmetic-progressions article covers the additive case in full.
Can the common ratio be negative or a fraction?
Yes to both. A negative ratio makes terms alternate in sign; a fractional ratio (∣r∣<1|r| < 1) makes the terms shrink toward zero, which is what allows an infinite GP to have a finite sum.
How do I find the common ratio if I only know two terms?
Divide the later term by the earlier one if they are consecutive. If they are not consecutive, set up am=arm−1a_m = ar^{m-1} and an=arn−1a_n = ar^{n-1} and divide the equations to isolate rr.
✍️ Written By
Madhukar V
Madhukar V
Math Columnist & Analyst
I see mathematics as more than just numbers and formulas—it is a way of thinking, questioning, and understanding the world around us. I work at the intersection of mathematics, EdTech, technology, and AI-powered learning, I enjoy exploring how mathematical ideas can be made simple, meaningful, and relatable. Through my articles, I aim to present mathematics with curiosity and a practical perspective, helping readers see the beauty and relevance behind every concept in a Bhanzu way.
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