Geometric progressions

Some sequences grow by adding the same amount each time; others grow by multiplying by the same amount each time — and multiplying, repeated, explodes far faster than adding ever could. A geometric progression (GP) is that second kind, a list like $3, 6, 12, 24, \dots$ where each term is the previous one times a fixed constant ratio. The same idea describes a population doubling every generation, a bouncing ball's height halving on each rebound, or money compounding at a steady percentage — which is why GPs sit right alongside arithmetic progressions in the algebra syllabus, as the multiplicative twin of the additive pattern you already know. This article builds the two formulas that do all the work for a GP — the general term $a_n$ and the sum $S_n$ of a finite number of terms — shows exactly where the sum formula comes from using a one-line trick, and works through ICSE-style problems: finding a specific term, finding a sum with $r>1$ , and finding a sum with a fractional $r<1$ . Along the way you will see the handful of slips (AP formulas creeping into GP questions, a wrong exponent, a division by zero) that cost the most marks.

The core insight

A geometric progression is a sequence with a constant common ratio $r$ between consecutive terms, so its $n$ th term is $a_n = ar^{n-1}$ and the sum of its first $n$ terms is $S_n = \dfrac{a(r^n-1)}{r-1}$ for $r \ne 1$ .

The core idea

A geometric progression is a sequence in which every term after the first is obtained by multiplying the term before it by a fixed number, the common ratio $r$ . If the first term is $a$ , the GP looks like

a,\ ar,\ ar^2,\ ar^3,\ \dots

Finding $r$ . Since consecutive terms are always in the same ratio, it is computed as (any term) divided by (the term just before it):

r = \dfrac{a_2}{a_1} = \dfrac{a_3}{a_2} = \dots

where: $a_1, a_2, a_3, \dots$ are the 1st, 2nd, 3rd, ... terms of the sequence. Checking a single pair of consecutive terms is not enough — a sequence can coincidentally match one ratio and fail the next — so a genuine GP needs every consecutive pair to give the same $r$ . That ratio can be a whole number, a fraction, or even negative; a negative $r$ produces a GP whose terms alternate in sign, such as $2, -6, 18, -54, \dots$ (here $r=-3$ ).

The general (nth) term. To reach the $n$ th term from the first term, you multiply by $r$ a total of $(n-1)$ times, not $n$ times, because the first term itself already counts as "zero multiplications" by $r$ :

a_n = ar^{n-1}

where: $a_n$ is the $n$ th term, $a$ is the first term, $r$ is the common ratio, and $n$ is a positive whole number (the position in the list).

The sum of the first $n$ terms (only defined this way for $r \ne 1$ ):

S_n = \dfrac{a(r^n-1)}{r-1} = \dfrac{a(1-r^n)}{1-r}

where: $S_n$ is the sum of the first $n$ terms; the two forms are algebraically identical (multiply top and bottom of either by $-1$ ), so use whichever keeps the denominator positive — the left form when $r>1$ , the right form when $r<1$ . If $r=1$ , every term equals $a$ and the sum is simply $S_n = na$ . The figure below shows why a GP grows so much faster than an AP: each new bar is the same multiple of the one before, not the same addition, so the bars shoot upward rather than climbing in equal steps.

Figure

A bar chart of the terms 3 6 12 24 48 of a geometric progression with each bar twice the height of the one before it showing multiplicative growth

The sum formula, from first principles

Derivation AWhere the geometric sum formula comes from

The goal is to derive $S_n = \dfrac{a(r^n-1)}{r-1}$ from nothing but the definition of a GP, using a shift-and-subtract trick that needs no calculus and works for any first term, common ratio (other than $r=1$ ), or number of terms.

Step 1. Write the sum $S_n$ of the first $n$ terms in full, then write $r$ times that same sum, term by term:

S_n = a + ar + ar^2 + \dots + ar^{n-1}

rS_n = ar + ar^2 + ar^3 + \dots + ar^{n}

Every term in the second line is simply the matching term of the first line, shifted one place and multiplied once more by $r$ .

Step 2. Subtract the first line from the second. Because the two lines share every middle term ( $ar, ar^2, \dots, ar^{n-1}$ ), those all cancel, leaving only the one term unique to each line:

rS_n - S_n = ar^n - a

Step 3. Factor both sides. On the left, $S_n$ is common; on the right, $a$ is common:

S_n(r-1) = a(r^n-1)

Step 4. Divide both sides by $(r-1)$ to isolate $S_n$ , valid only when $r \ne 1$ since dividing by zero is undefined:

S_n = \dfrac{a(r^n-1)}{r-1}

Check: for $a=3$ , $r=2$ , $n=4$ (the GP $3,6,12,24$ ), $S_4 = \dfrac{3(2^4-1)}{2-1} = \dfrac{3(15)}{1}=45$ , matching $3+6+12+24=45$ by direct addition.

Worked example

Question: Find the 7th term and the sum of the first 7 terms of the GP $3, 6, 12, 24, \dots$

Step 1. Identify $a$ and $r$ . Here $a=3$ and $r = \dfrac{6}{3}=2$ .

Step 2. Find the 7th term using $a_n = ar^{n-1}$ :

a_7 = 3\times2^{7-1} = 3\times2^6 = 3\times64 = 192

Step 3. Find the sum of the first 7 terms using $S_n = \dfrac{a(r^n-1)}{r-1}$ , choosing this form since $r=2>1$ :

S_7 = \dfrac{3(2^7-1)}{2-1} = \dfrac{3(128-1)}{1} = 3\times127 = 381

So $a_7 = 192$ and $S_7 = 381$ .

Competitive stretch

The first term of a GP is $16$ and the common ratio is $\dfrac{1}{2}$ . Find the sum of the first 5 terms, and verify it by listing the terms directly.

Step 1. Here $a=16$ , $r=\dfrac12$ , $n=5$ . Since $r<1$ , use the second form $S_n = \dfrac{a(1-r^n)}{1-r}$ to keep the denominator positive:

S_5 = \dfrac{16\left(1-\left(\dfrac12\right)^5\right)}{1-\dfrac12} = \dfrac{16\left(1-\dfrac{1}{32}\right)}{\dfrac12} = 16\times\dfrac{31}{32}\times2 = \dfrac{992}{32} = 31

Step 2. Verify by listing terms directly: the GP is $16, 8, 4, 2, 1$ , and $16+8+4+2+1=31$ , which matches. So $S_5 = 31$ . Listing terms is a good habit whenever $n$ is small — it catches formula-substitution slips, like using the wrong exponent, that might otherwise slip through unnoticed; for larger $n$ it becomes impractical and the formula is essential.

Common traps

Using the AP sum/term formulas by mistake on a GP, or vice versa — always check whether consecutive terms share a common difference (AP) or a common ratio (GP) before choosing a formula.
Writing $r^n$ instead of $r^{n-1}$ in the general term — the first term itself corresponds to zero powers of $r$ , so the exponent is always one less than the term's position.
Applying $S_n = \dfrac{a(r^n-1)}{r-1}$ when $r=1$ — this formula divides by zero at $r=1$ ; a GP with $r=1$ is just a constant sequence, and $S_n=na$ directly in that special case.
Confusing a finite GP sum with the (out-of-scope) infinite GP sum — at this level, only sums of a specified finite number of terms are required.

Ask the tutor: want to practise identifying whether a sequence is an AP or a GP, or check your own sum calculation step by step — ask the AI tutor to walk through your specific sequence.

Why does the common ratio sometimes come out negative?

A GP only requires a constant ratio between consecutive terms, and nothing forbids that ratio from being negative. In $2, -6, 18, -54, \dots$ each term is $-3$ times the one before, so $r=-3$ , and the terms alternate between positive and negative as a result. The formulas $a_n=ar^{n-1}$ and $S_n=\dfrac{a(r^n-1)}{r-1}$ still apply exactly as written; you simply get an alternating sign pattern when you compute powers of a negative $r$ , so double-check the sign of $r^{n-1}$ carefully.

How do I quickly tell whether a given sequence is an AP or a GP?

Test both operations on the first few terms. Subtract consecutive terms: if that difference stays constant all the way through, it is an AP. Divide consecutive terms instead: if that ratio stays constant, it is a GP. Do not stop after checking just one pair — a sequence can coincidentally match for one gap or ratio and fail the next, so confirm the pattern holds across at least three consecutive pairs before committing to a formula.

Do I ever need the sum of an infinite GP at this level?

No. ICSE Class 10 covers only the sum of a specified finite number of terms of a GP, using $S_n=\dfrac{a(r^n-1)}{r-1}$ . The formula for summing an infinite GP (which only converges when $|r|<1$ ) belongs to a later stage of study and is out of scope here — if a question gives you a definite value of $n$ , it wants the finite-sum formula, not an infinite one.