What Is The Sum Of A GP?
The sum of a GP is the result of adding the terms of a geometric progression - a sequence where each term is the previous term multiplied by a fixed number called the common ratio. If the progression is $a, ar, ar^2, ar^3, \dots$, then the sum of its first $n$ terms, written $S_n$, is the total of those $n$ terms added together.
A quick definition before the formula. A geometric progression (or geometric sequence) is an ordered list like $3, 6, 12, 24, \dots$ where you multiply by the same number each step - here, $2$. That multiplier is the common ratio, $r$. The first number is the first term, $a$.
What Is The Formula For The Sum Of n Terms Of A GP?
There are two cases, and choosing the right one is the whole game.
When $r \neq 1$:
$$S_n = \frac{a(1 - r^n)}{1 - r}$$
An equivalent form, handy when $r > 1$ so you work with positive numbers:
$$S_n = \frac{a(r^n - 1)}{r - 1}$$
The two are identical; multiplying the top and bottom of one by $-1$ gives the other. Use whichever keeps the arithmetic cleaner.
When $r = 1$:
$$S_n = na$$
Variable glossary:
Symbol | Meaning |
|---|---|
$a$ | the first term of the progression |
$r$ | the common ratio (each term ÷ the one before) |
$n$ | the number of terms being added |
$S_n$ | the sum of those first $n$ terms |
Where Does The Formula Come From?
This is the part worth seeing once. The derivation is a two-line trick called multiply-and-subtract, and it shows up again in calculus and in solving recurrences.
Write the sum out in full:
$$S_n = a + ar + ar^2 + \dots + ar^{n-1}$$
Now multiply every term by $r$:
$$r S_n = ar + ar^2 + ar^3 + \dots + ar^n$$
Line the two equations up and subtract the second from the first. Almost everything in the middle cancels - the $ar$ in the first matches the $ar$ in the second, the $ar^2$ matches the $ar^2$, and so on. Only the very first term of $S_n$ and the very last term of $rS_n$ survive:
$$S_n - r S_n = a - ar^n$$
Factor both sides:
$$S_n(1 - r) = a(1 - r^n)$$
Divide by $(1 - r)$:
$$S_n = \frac{a(1 - r^n)}{1 - r}$$
Now the $r = 1$ case explains itself. The last step divided by $(1 - r)$, and if $r = 1$ that is division by zero, so the formula simply does not apply. Go back to the start: when $r = 1$ every term equals $a$, so adding $n$ of them gives $na$. The separate case is not an exception bolted on; it is the one place the derivation cannot reach.
Examples Of The Sum Of A GP
Example 1
Find the sum of the first 5 terms of $1, 2, 4, 8, 16$.
Here $a = 1$, $r = 2$, $n = 5$. Since $r > 1$, use the $r^n - 1$ form:
$$S_5 = \frac{1(2^5 - 1)}{2 - 1}$$
$$S_5 = \frac{32 - 1}{1}$$
$$S_5 = 31$$
Final answer: $S_5 = 31$. Check by adding directly: $1 + 2 + 4 + 8 + 16 = 31$.
Example 2
Find the sum of the first 6 terms of $4, 8, 16, \dots$ where every term is the same... or is it?
Suppose a student misreads the progression as having "no growth" and reaches for $S_n = na$, computing $S_6 = 6 \times 4 = 24$. Take a second: the terms $4, 8, 16$ are clearly not all equal, so $r$ is not $1$ - it is $2$, and $24$ is far too small (the sixth term alone is $128$). The rule $S_n = na$ applies only when every term is identical. Switch to the correct formula:
$$S_6 = \frac{4(2^6 - 1)}{2 - 1}$$
$$S_6 = \frac{4(64 - 1)}{1}$$
$$S_6 = 4 \times 63 = 252$$
Final answer: $S_6 = 252$. The lesson: confirm what $r$ actually is before choosing between $S_n = na$ and the ratio formula.
Example 3
Find the sum of the first 8 terms of $1, \frac{1}{2}, \frac{1}{4}, \dots$.
Now $a = 1$, $r = \frac{1}{2}$, $n = 8$. Because $r < 1$, the $1 - r^n$ form keeps the signs tidy:
$$S_8 = \frac{1\left(1 - \left(\frac{1}{2}\right)^8\right)}{1 - \frac{1}{2}}$$
$$S_8 = \frac{1 - \frac{1}{256}}{\frac{1}{2}}$$
$$S_8 = \frac{\frac{255}{256}}{\frac{1}{2}}$$
$$S_8 = \frac{255}{256} \times 2 = \frac{255}{128}$$
Final answer: $S_8 = \dfrac{255}{128}$, which is $1.9921875$. Notice it is creeping toward $2$ but has not reached it - a preview of what happens when the terms run forever, covered in the sum of an infinite GP.
Example 4
A GP has first term $5$ and common ratio $3$. Find the sum of the first 4 terms.
$a = 5$, $r = 3$, $n = 4$:
$$S_4 = \frac{5(3^4 - 1)}{3 - 1}$$
$$S_4 = \frac{5(81 - 1)}{2}$$
$$S_4 = \frac{5 \times 80}{2} = \frac{400}{2} = 200$$
Final answer: $S_4 = 200$. Direct check: $5 + 15 + 45 + 135 = 200$.
Example 5
A GP with negative ratio: find the sum of the first 5 terms of $2, -6, 18, -54, \dots$.
The ratio is negative: $r = \frac{-6}{2} = -3$. Sign care matters here.
$$S_5 = \frac{2\left((-3)^5 - 1\right)}{-3 - 1}$$
$$S_5 = \frac{2(-243 - 1)}{-4}$$
$$S_5 = \frac{2 \times (-244)}{-4}$$
$$S_5 = \frac{-488}{-4} = 122$$
Final answer: $S_5 = 122$. Check: $2 - 6 + 18 - 54 + 162 = 122$. The most common error in this kind of problem is computing $(-3)^5$ as $243$ instead of $-243$; an odd power of a negative number stays negative.
Example 6
A folding-paper problem. A sheet of paper is $0.1$ mm thick. You fold it in half, then in half again, repeatedly. After each fold the stack doubles. What is the total of all the layer-thicknesses added across folds $1$ through $7$?
The thickness after fold $k$ forms a GP with $a = 0.2$ mm (the stack after the first fold), $r = 2$, and $n = 7$:
$$S_7 = \frac{0.2(2^7 - 1)}{2 - 1}$$
$$S_7 = \frac{0.2(128 - 1)}{1}$$
$$S_7 = 0.2 \times 127 = 25.4 \text{ mm}$$
Final answer: $S_7 = 25.4$ mm. Geometric growth is deceptively fast, which is the whole point of the rumour story at the top.
Why The Multiply-And-Subtract Trick Matters Beyond GPs
"Each step doubles, so adding fixed numbers will never describe it."
The reason a separate sum formula exists for geometric progressions, rather than the steady-step formula you use for an arithmetic sequence, is that geometric growth is multiplicative. A population that grows by a fixed percentage each year, money compounding in an account, a signal halving as it passes through layers of shielding - all of these are GPs, and the running total of any of them is a sum of a GP.
The historical roots run deep. The puzzle of the king who promised a grain of rice on the first square of a chessboard, two on the second, four on the third, and so on, is a sum of a GP with $a = 1$, $r = 2$, $n = 64$. The total - over eighteen quintillion grains - is more rice than has ever been grown. That story survives precisely because the sum of a GP defies the intuition that "adding things up" stays manageable.
The multiply-and-subtract move from the derivation is worth keeping. You will meet it again when summing series in calculus and when collapsing recurrences. It is one of those techniques that pays rent across many topics, which is part of why teaching the derivation matters more than handing over the result.
Common Errors That Cost Marks
Mistake 1: Forgetting to check whether r = 1
Where it slips in: Reaching for $S_n = \frac{a(1 - r^n)}{1 - r}$ on autopilot without glancing at the ratio.
Don't do this: Apply the ratio formula when $r = 1$, producing a $\frac{0}{0}$ and a stalled solution.
The correct way: Identify $a$, $r$, and $n$ first. If $r = 1$, every term equals $a$, so $S_n = na$. Naming $r$ before picking a formula is the habit that prevents the most common stall in these problems, because the two formulas are not interchangeable.
Mistake 2: Using the wrong power on r
Where it slips in: Writing $S_n = \frac{a(1 - r)}{1 - r}$ or using $r^{n-1}$ where $r^n$ belongs.
Don't do this: Confuse the $n$th-term formula $a_n = ar^{n-1}$ with the sum formula. The term formula has the $n - 1$ power; the sum formula has the $n$ power.
The correct way: The sum formula carries $r^n$, full stop. The student who memorised both formulas but never sees them side by side is the one who swaps the exponents. Write them next to each other once and the difference sticks: $a_n = ar^{n-1}$, $S_n = \frac{a(1-r^n)}{1-r}$.
Mistake 3: Sign slips when r is negative
Where it slips in: Progressions like $2, -6, 18, \dots$ where odd and even powers of $r$ flip sign.
Don't do this: Treat $(-3)^5$ as positive, or forget that the denominator $1 - r$ becomes $1 - (-3) = 4$, not $1 - 3$.
The correct way: Substitute the negative $r$ with brackets around it - $(-3)$, not $-3$ - and evaluate the power carefully. Odd powers of a negative stay negative; even powers turn positive.
Practice Questions On The Sum Of A GP
Identify $a$, $r$, and $n$ before choosing a formula. Answers follow.
Find the sum of the first 6 terms of $2, 4, 8, \dots$.
A GP has first term $3$ and common ratio $1$. Find the sum of its first 5 terms.
Find the sum of the first 4 terms of $1, \frac{1}{3}, \frac{1}{9}, \dots$.
Find the sum of the first 5 terms of $2, -4, 8, -16, \dots$.
Find the sum of the first 10 terms of a GP with $a = 5$ and $r = 2$.
Answers:
$a = 2$, $r = 2$, $n = 6$: $S_6 = \dfrac{2(2^6 - 1)}{2 - 1} = 126$.
$r = 1$, so $S_5 = na = 5 \times 3 = 15$.
$a = 1$, $r = \frac{1}{3}$, $n = 4$: $S_4 = \dfrac{1\left(1 - \left(\frac{1}{3}\right)^4\right)}{1 - \frac{1}{3}} = \dfrac{40}{27}$.
$a = 2$, $r = -2$, $n = 5$: $S_5 = \dfrac{2\left((-2)^5 - 1\right)}{-2 - 1} = \dfrac{2(-33)}{-3} = 22$.
$a = 5$, $r = 2$, $n = 10$: $S_{10} = \dfrac{5(2^{10} - 1)}{2 - 1} = 5 \times 1023 = 5115$.
The Short Version
The sum of a GP for the first $n$ terms is $S_n = \frac{a(1 - r^n)}{1 - r}$ when $r \neq 1$.
When $r = 1$, every term is equal, so $S_n = na$.
The formula comes from multiplying the sum by $r$ and subtracting, which cancels the middle terms.
Identify $a$, $r$, and $n$ before choosing a formula; sign care matters when $r$ is negative.
Geometric sums grow far faster than intuition suggests, which is why they model rumours, compounding, and chain reactions.
Try These Next
Now compute the sum of the first $6$ terms of $3, 6, 12, \dots$ on your own, then redo it by direct addition to confirm. If the derivation felt fast, rewrite the multiply-and-subtract steps from a blank page until the cancellation is obvious - that single technique is worth more than the memorised result.
At Bhanzu, our trainers teach the GP sum by deriving it live rather than posting the formula on a slide, so students own the cancellation trick and can rebuild the formula if they forget it. Want a live trainer to work through more progression problems? Book a free demo class.
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