What is the nth Term of a GP?
The nth term of a GP is the value of the term in position $n$ of a geometric progression, given by the formula $a_n = ar^{n-1}$. 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 $r$. In $3, 6, 12, 24, \dots$ each term is double the last, so $r = 2$ and the first term $a = 3$. You find $r$ by dividing any term by the one before it:
$$r = \frac{a_2}{a_1} = \frac{6}{3} = 2$$
A general GP is written:
$$a, ; ar, ; ar^2, ; ar^3, ; \dots$$
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:
$$a_n = ar^{n-1}$$
where $a$ is the first term, $r$ is the common ratio, $n$ is the position of the term you want, and $a_n$ is its value.
The derivation is just counting how many times you multiply by $r$. To reach the first term you multiply by $r$ zero times; the second term, once; the $n$th term, $(n-1)$ times:
$a_1 = a$
$a_2 = a \cdot r = ar^1$
$a_3 = ar \cdot r = ar^2$
$a_n = ar^{n-1}$
The exponent is $(n-1)$, not $n$, 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 $n$ terms and last term $l$, the $k$th term from the end is:
$$a_{n-k+1} = l \cdot \left(\frac{1}{r}\right)^{k-1}$$
You divide by $r$ as you walk backwards, which is the same as multiplying by $\dfrac{1}{r}$ — 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 $a_n = ar^{n-1}$.
Example 1
Find the common ratio of $5, 15, 45, 135, \dots$
Divide a term by the one before it: $\dfrac{15}{5} = 3$ and $\dfrac{45}{15} = 3$.
Final answer: $r = 3$.
Example 2
Find the 6th term of $2, 6, 18, 54, \dots$
A frequent first move is to write $a_6 = ar^n = 2 \cdot 3^6$.
Why that is wrong. Using $r^n$ instead of $r^{n-1}$ 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 $r$ only five times.
The correct method. Here $a = 2$, $r = 3$, $n = 6$:
$a_6 = ar^{n-1} = 2 \cdot 3^{6-1} = 2 \cdot 3^5$
$a_6 = 2 \cdot 243 = 486$
Final answer: $486$. The exponent is $n-1$, every time.
Example 3
Find the 8th term of $1, 2, 4, 8, \dots$
Here $a = 1$, $r = 2$, $n = 8$:
$a_8 = 1 \cdot 2^{8-1} = 2^7 = 128$
Final answer: $128$.
Example 4
Find the 5th term of a GP with first term $2$ and common ratio $\dfrac{1}{3}$.
Here $a = 2$, $r = \dfrac{1}{3}$, $n = 5$:
$a_5 = 2 \cdot \left(\frac{1}{3}\right)^{5-1} = 2 \cdot \left(\frac{1}{3}\right)^{4}$
$a_5 = 2 \cdot \frac{1}{81} = \frac{2}{81}$
Final answer: $\dfrac{2}{81}$. When $|r| < 1$ the terms shrink, which is what powers the infinite-sum case.
Example 5
In the GP $3, 6, 12, \dots$, which term equals $192$?
Here $a = 3$, $r = 2$, $a_n = 192$. Solve for $n$:
$192 = 3 \cdot 2^{n-1}$
$64 = 2^{n-1}$
$2^6 = 2^{n-1}$
$n - 1 = 6$, so $n = 7$.
Final answer: the 7th term.
Example 6
A bacterial culture starts with $500$ cells and triples every hour. How many cells after $4$ hours?
The hourly counts form a GP: $a = 500$, $r = 3$. After $4$ hours is the 5th term ($n = 5$, since the start counts as term 1):
$a_5 = 500 \cdot 3^{5-1} = 500 \cdot 3^4$
$a_5 = 500 \cdot 81 = 40500$
Final answer: $40{,}500$ 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 $n$ 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 $ar^{n-1}$ is the only way to see exactly how fast.
Common Mistakes With the Nth Term of a GP
Mistake 1: Using $r^n$ instead of $r^{n-1}$
Where it slips in: Computing a specific term. The rusher reads "6th term" and raises $r$ to the 6th power.
Don't do this: $a_6 = ar^6$.
The correct way: $a_6 = ar^5$. 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 $(n-1)$ 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 $4, 12, 36, \dots$ write $r = 12 - 4 = 8$.
The correct way: $r = \dfrac{12}{4} = 3$. The common ratio comes from division, not subtraction. If dividing consecutive terms gives the same number each time, that number is $r$; 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 $r$ 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 $\left(\frac{1}{3}\right)^4$ as $\frac{1}{3^{}} = \frac{1}{3}$, powering only the bottom partially, or write $(-2)^3$ as $2^3$.
The correct way: Power the whole ratio: $\left(\frac{1}{3}\right)^4 = \dfrac{1}{81}$, and keep the sign inside the bracket: $(-2)^3 = -8$. A negative $r$ 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 $a_n = ar^{n-1}$ — first term times the ratio raised to one less than the position.
The exponent is $(n-1)$, not $n$, because the first term exists before any multiplication.
The common ratio $r$ 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.
Find the 7th term of $3, 6, 12, 24, \dots$ (Revisit Example 2 and Mistake 1.)
Find the common ratio of $81, 27, 9, 3, \dots$ (Revisit Example 1 and Mistake 2.)
Which term of $2, 8, 32, \dots$ equals $512$? (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.
Read More
Geometric progressions — the full GP overview, including properties.
Arithmetic sequence — the add-a-difference counterpart.
Sigma notation — compact notation for the sums a GP produces.
Exponents — the powers of $r$ that drive the nth-term formula.
Progression in maths — how geometric progressions sit alongside arithmetic ones.
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