Geometric progressions

Some sequences grow by addition, some by multiplication — and the multiplicative kind turns up everywhere from compound interest to bacterial colonies to the bounce of a ball. A Geometric Progression (GP) is exactly that: a sequence where each term is the previous term times a fixed number, the common ratio $r$ . In CBSE Class 11's Sequence and Series chapter, GP is the natural partner to the Arithmetic Progression you already know, and it is fully examined — the general term, the sum of a finite number of terms, and the sum of infinitely many terms (when that sum even exists) all appear in board and competitive papers. This article compresses the GP toolkit into one place: how the formulas are built, where the $|r| < 1$ condition for an infinite sum comes from, and the traps that cost the most marks.

The core insight

In a GP each term equals the previous term times a fixed common ratio $r$ , so the $n$ -th term is $a_n = ar^{n-1}$ , the sum of the first $n$ terms is $S_n = \dfrac{a(r^n-1)}{r-1}$ , and an infinite GP has a finite sum $S_\infty = \dfrac{a}{1-r}$ only when $|r| < 1$ .

The core idea

A sequence $a_1, a_2, a_3, \dots$ is a Geometric Progression if the ratio of every term to the one before it is the same fixed, non-zero number $r$ :

\frac{a_{n+1}}{a_n} = r \quad \text{for all } n \ge 1

where: $r$ is the common ratio and $a_1$ (often just written $a$ ) is the first term. Multiplying the first term by $r$ repeatedly generates the whole sequence, $a,\ ar,\ ar^2,\ ar^3,\dots$ , which immediately gives the general term.

General term. Since each step multiplies by $r$ once more, the $n$ -th term is

a_n = a\,r^{\,n-1}

where: $a$ is the first term, $r$ is the common ratio, and $n$ is the position of the term counted from $1$ . This single formula lets you jump to any term without listing every one before it.

Sum of the first $n$ terms. For $r \neq 1$ ,

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

where: the two forms are algebraically identical (both top and bottom are negated together); pick whichever avoids a negative denominator for the $r$ you're given. If $r = 1$ , every term equals $a$ , so $S_n = na$ trivially.

Sum to infinity. When $|r| < 1$ , successive terms $ar^{n-1}$ shrink toward zero, and the partial sums settle at a finite limit:

S_\infty = \frac{a}{1-r}, \qquad |r| < 1

If $|r| \ge 1$ , the terms don't shrink (they stay the same size or grow), so the sum keeps changing without approaching a fixed value — $S_\infty$ simply does not exist as a finite number in that case.

Recognizing a GP quickly. Given a list of numbers, don't guess — divide consecutive terms. If $a_2/a_1$ , $a_3/a_2$ , and $a_4/a_3$ all come out equal, you have a GP and that shared value is $r$ ; if even one ratio differs, it isn't a GP no matter how "regular" the numbers look. This is the same discipline that separates a GP from an AP: an AP has a constant difference, a GP has a constant ratio, and mixing the two tests up is the single most common identification error at this level.

Figure

Bar heights for the terms of a doubling geometric progression starting at three, showing each bar twice the height of the one before it

Derivation ADeriving the sum of a finite GP

The goal is to derive the closed-form formula $S_n = \dfrac{a(r^n-1)}{r-1}$ for the sum of the first $n$ terms of a GP with first term $a$ and common ratio $r \neq 1$ , starting from nothing but writing out the sum term by term and a single clever subtraction, with no formula assumed in advance.

Step 1. Write the sum out fully. Let $S_n = a + ar + ar^2 + \cdots + ar^{n-1}$ , the sum of the first $n$ terms.

Step 2. Multiply the whole equation by $r$ . This shifts every term one place along: $rS_n = ar + ar^2 + ar^3 + \cdots + ar^{n}$ .

Step 3. Subtract the second equation from the first. Almost everything cancels because the two sums share the terms $ar, ar^2, \dots, ar^{n-1}$ : $S_n - rS_n = a - ar^n$ , so $S_n(1-r) = a(1-r^n)$ .

Step 4. Divide both sides by $(1-r)$ , which is safe precisely because $r \neq 1$ : $S_n = \dfrac{a(1-r^n)}{1-r}$ , which is the same expression as $\dfrac{a(r^n-1)}{r-1}$ after multiplying top and bottom by $-1$ .

Step 5. Sanity-check with a tiny case. Take $a = 1$ , $r = 2$ , $n = 3$ , so the sum should be $1+2+4=7$ . The formula gives $S_3 = \dfrac{1(2^3-1)}{2-1} = \dfrac{7}{1} = 7$ — matching the direct addition and confirming the derivation, and this same shift-and-subtract trick is exactly what later produces $S_\infty = \dfrac{a}{1-r}$ once $r^n \to 0$ for $|r|<1$ .

Worked example

Problem. 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 $r = 6/3 = 2$ (check: $12/6 = 2$ and $24/12 = 2$ , so the ratio really is constant).

Step 2. Find the 7th term using $a_n = ar^{n-1}$ : $a_7 = 3 \times 2^{7-1} = 3 \times 64 = 192$ .

Step 3. Find $S_7$ using $S_n = \dfrac{a(r^n-1)}{r-1}$ : $S_7 = \dfrac{3(2^7-1)}{2-1} = 3 \times 127 = 381$ .

The 7th term is $192$ , and the sum of the first 7 terms is $\mathbf{381}$ (direct check: $3+6+12+24+48+96+192 = 381$ ).

Competitive stretch

Problem. Find the sum to infinity of the GP $8, -4, 2, -1, \dots$

Step 1. Identify $a$ and $r$ : $a = 8$ , $r = -4/8 = -\tfrac12$ (check: $2/(-4) = -\tfrac12$ , $-1/2 = -\tfrac12$ ).

Step 2. Confirm $S_\infty$ exists: $|r| = \tfrac12 < 1$ , so it does.

Step 3. Apply $S_\infty = \dfrac{a}{1-r} = \dfrac{8}{1-(-\tfrac12)} = \dfrac{8}{3/2} = \dfrac{16}{3}$ . The sum to infinity is $\mathbf{16/3}$ — the alternating sign of $r$ doesn't stop convergence, since only $|r| < 1$ matters, not the sign. This is exactly the kind of reasoning that carries over to compound-interest and depreciation problems outside the textbook: a quantity repeatedly scaled by a fixed factor smaller than one — money losing a fixed percentage each year, or a bouncing ball losing height each bounce — is a GP with $|r|<1$ , so its "total over all time" is a genuine finite number given by $S_\infty = \dfrac{a}{1-r}$ , not an ever-growing quantity.

Common traps

Using $S_\infty = \dfrac{a}{1-r}$ without checking $|r| < 1$ first. If $|r| \ge 1$ , the sum to infinity does not exist and the formula is meaningless.
Sign errors switching between $\dfrac{a(r^n-1)}{r-1}$ and $\dfrac{a(1-r^n)}{1-r}$ . They are equal, but only if both numerator and denominator are negated together — negating just one flips the answer's sign.
Confusing common ratio with common difference. A GP is defined by a constant ratio $a_{n+1}/a_n$ , not a constant difference; testing "AP or GP" means checking division, not subtraction.
Applying GP formulas to a non-geometric sequence. Always verify $a_{n+1}/a_n$ is the same for every consecutive pair before using $a_n = ar^{n-1}$ , $S_n$ , or $S_\infty$ .

Ask the tutor: Want to try spotting a GP hidden inside a word problem, or practise deciding when a sum to infinity is actually defined?

How is a GP different from an AP, in one line?

An AP adds the same fixed number to get the next term ( $a_{n+1} = a_n + d$ ); a GP multiplies by the same fixed number instead ( $a_{n+1} = r\,a_n$ ). That single difference — addition versus multiplication — is why AP terms grow at a steady rate while GP terms can shoot up (or shrink toward zero) very fast, and why the two chapters use completely different-looking formulas for their sums.

Why does the sum to infinity need the common ratio's absolute value less than 1 — what actually goes wrong otherwise?

The infinite sum is really the limit of $S_n = \dfrac{a(1-r^n)}{1-r}$ as $n \to \infty$ . That limit exists only if $r^n \to 0$ , which happens precisely when $|r| < 1$ — repeated multiplication by a fraction smaller than one in size shrinks the term toward zero. If $|r| \ge 1$ , $r^n$ either stays the same size, grows, or oscillates without settling, so $S_n$ never approaches one fixed number, and "the sum to infinity" simply isn't a real number in that case.

If a problem gives three numbers and asks whether they form a GP, what should I check?

Compute the ratio of the second to the first, and the ratio of the third to the second. If both ratios come out equal, the three numbers form a GP with that common ratio; if not, they don't. This is the same test used for any longer sequence — a GP is defined entirely by having a constant ratio between every pair of consecutive terms, so checking just one matching pair is not enough if more terms are given.

Geometric progressions

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads