A Pattern That Doubles, Triples, or Halves Forever
Some sequences add a fixed amount each step. Others multiply by a fixed amount. The second kind grows — or shrinks — much faster, and that single difference reshapes how the sequence behaves.
The geometric progression 2, 4, 8, 16, … reaches 256 by term 8. The arithmetic progression 2, 4, 6, 8, … reaches 16. Same start, same step count, sixteen times the difference. That gap is why compound interest beats simple interest, why pandemics start slow and then aren't, and why the 1859 release of 24 rabbits in Victoria became 10 billion rabbits across Australia.
What a Geometric Progression Is
A geometric progression (GP), also called a geometric sequence, is a list of numbers $a_1, a_2, a_3, \ldots$ such that the ratio between any two consecutive terms is the same constant $r$:
$$\frac{a_2}{a_1} = \frac{a_3}{a_2} = \frac{a_4}{a_3} = \cdots = r.$$
$r$ is the common ratio. The first term is $a$ (sometimes $a_1$). The general term is
$$a_n = a \cdot r^{n-1}.$$
Examples — $2, 6, 18, 54, \ldots$ ($a = 2$, $r = 3$). $;;10, 5, 2.5, 1.25, \ldots$ ($a = 10$, $r = 1/2$). $;;1, -3, 9, -27, \ldots$ ($a = 1$, $r = -3$).
Quick facts.
General term: $a_n = a \cdot r^{n-1}$.
Common ratio: $r = a_{n+1} / a_n$, constant for all $n$.
Finite sum: $S_n = a(1 - r^n) / (1 - r)$ for $r \neq 1$.
Infinite sum: $S_{\infty} = a / (1 - r)$ when $|r| < 1$; diverges when $|r| \geq 1$.
Growth behaviour: exponential if $|r| > 1$, decay if $0 < r < 1$, oscillation if $r < 0$.
Grade introduced: CBSE Class 11; CCSS-M HSF-LE.A.2 (construct linear and exponential functions, including geometric sequences); NCERT Class 11 Chapter 9 — Sequences and Series.
The nth Term Formula
To find any specific term of a GP, multiply the first term by $r$ as many times as needed.
$$a_n = a \cdot r^{n-1}.$$
The exponent is $n - 1$ because the first term involves zero multiplications: $a_1 = a \cdot r^0 = a$.
Example. Find the 7th term of $3, 6, 12, 24, \ldots$. Here $a = 3$, $r = 2$, $n = 7$.
$$a_7 = 3 \cdot 2^{6} = 3 \cdot 64 = 192.$$
The Sum of a Finite GP
The sum of the first $n$ terms is
$$S_n = \frac{a(1 - r^n)}{1 - r}, \quad r \neq 1.$$
When $r = 1$ every term equals $a$ and $S_n = na$.
Derivation. Let $S_n = a + ar + ar^2 + \cdots + ar^{n-1}$. Multiply by $r$: $rS_n = ar + ar^2 + \cdots + ar^n$. Subtract: $S_n - rS_n = a - ar^n$, so $S_n(1 - r) = a(1 - r^n)$. Divide by $1 - r$.
The derivation takes three lines. The formula it produces holds for every $r \neq 1$.
The Sum of an Infinite GP — When It Exists
When $|r| < 1$, the powers $r^n$ shrink to zero as $n \to \infty$. The finite sum formula becomes
$$S_{\infty} = \frac{a}{1 - r}, \quad |r| < 1.$$
When $|r| \geq 1$, the terms do not shrink — they grow or stay the same size — and the sum diverges.
Example. $1 + \tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{8} + \cdots = \tfrac{1}{1 - 1/2} = 2$. The infinite sum equals 2 exactly — Zeno's paradox, resolved.
Three Worked Examples of Geometric Progression
Quick. Find the 5th term of the GP $2, 6, 18, 54, \ldots$.
$a = 2$, $r = 3$, $n = 5$.
$$a_5 = 2 \cdot 3^{4} = 2 \cdot 81 = 162.$$
Final answer: $a_5 = 162$.
Standard (Wrong Path First — Where Things Go Sideways). Find the sum of the first 8 terms of $5, 10, 20, 40, \ldots$.
The wrong path. The rusher uses arithmetic-sequence muscle memory: $S = (n/2)(a_1 + a_n)$. They compute $a_8 = 5 \cdot 2^7 = 640$, then $S = (8/2)(5 + 640) = 4 \cdot 645 = 2580$.
Check by listing: $5 + 10 + 20 + 40 + 80 + 160 + 320 + 640 = 1275$. The wrong-formula answer is roughly double the true sum.
The flaw: the arithmetic-sum formula assumes terms grow linearly. GP terms grow exponentially. The early terms are tiny compared to the last — so an "average of first and last times count" massively overcounts.
The rescue. Use the GP sum formula: $S_n = a(1 - r^n)/(1 - r) = a(r^n - 1)/(r - 1)$ when $r > 1$.
$$S_8 = \frac{5(2^8 - 1)}{2 - 1} = \frac{5 \cdot 255}{1} = 1275.$$
Final answer: $S_8 = 1275$.
In Bhanzu's Grade 11 cohort, mistaking the arithmetic-sum formula for the GP-sum formula is the single most frequent error on the first sequences-and-series worksheet — roughly four in ten attempts. The trainer's fix is to write both sum formulas at the top of the board with one labelled "AP" and one labelled "GP," then have the student name which to use before plugging in numbers.
Stretch. A ball is dropped from 32 metres. Each bounce reaches $3/4$ of the previous height. What is the total vertical distance the ball travels before it stops bouncing?
Each bounce contributes up and down — except the initial drop, which is only down. The downward distances form a GP: $32 + 24 + 18 + 13.5 + \cdots$ with $r = 3/4$. The upward distances form the same GP starting from the first bounce: $24 + 18 + 13.5 + \cdots$.
Sum of downward distances: $S_{\downarrow} = 32 / (1 - 3/4) = 32 / (1/4) = 128$ m.
Sum of upward distances: $S_{\uparrow} = 24 / (1 - 3/4) = 24 / (1/4) = 96$ m.
Total: $128 + 96 = 224$ metres.
Final answer: the ball travels $224$ metres before coming to rest.
Where Geometric Progressions Show Up in the Real World
The GP is one of the most common patterns in nature and economics.
Compound interest. A principal $P$ at rate $r$ compounded annually for $n$ years grows as $P(1 + r)^n$ — a GP.
Population growth (early-stage). Before food or space limits, populations multiply by a constant factor each generation.
Radioactive decay. Half-life is the time for the GP $A_0, A_0/2, A_0/4, \ldots$ to shrink by half. After $n$ half-lives the amount is $A_0 / 2^n$.
Music intervals. The frequencies of equal-tempered semitones form a GP with common ratio $2^{1/12} \approx 1.05946$.
Pixel scaling. Doubling image resolution doubles the side length and quadruples pixel count — a GP in both directions.
Bacterial growth. E. coli doubles every 20 minutes under ideal conditions, producing the GP that lets a single cell become $2^{72}$ cells in 24 hours.
The destination, in every direction, is the same: any time something multiplies by a fixed factor with each step, the geometric progression is the model.
Geometric Progression: Mistakes Students Make Most Often
1. Confusing the common ratio with the common difference.
Where it slips in: A student writes $a_n = a + (n - 1)r$ — the arithmetic formula — for a GP.
Don't do this: Add the common ratio instead of multiplying.
The correct way: GP uses $a_n = a \cdot r^{n-1}$ — multiplication, not addition.
2. Using the wrong sum formula.
Where it slips in: Asked to sum a GP, the student uses the AP sum formula $S_n = (n/2)(a + a_n)$.
Don't do this: Apply $S = (n/2)(\text{first} + \text{last})$ to exponentially-growing terms.
The correct way: $S_n = a(1 - r^n)/(1 - r)$ for finite GP; $a/(1 - r)$ for infinite when $|r| < 1$.
3. Forgetting the $r = 1$ exception.
Where it slips in: Dividing by $1 - r$ when $r = 1$ is dividing by zero.
Don't do this: Plug $r = 1$ into the formula.
The correct way: Handle $r = 1$ separately. Every term equals $a$, so $S_n = na$.
4. Computing infinite sums when $|r| \geq 1$.
Where it slips in: Applying $S_\infty = a/(1-r)$ to a divergent series like $1 + 2 + 4 + 8 + \cdots$.
Don't do this: Treat $1/(1 - 2) = -1$ as the "sum" of a divergent positive series.
The correct way: Check $|r| < 1$ first. If not, the infinite sum does not exist.
The real-world version. In October 1859, Thomas Austin released 24 European rabbits on his property near Geelong, Victoria, Australia, intending to hunt them. Rabbits breed roughly six litters per year, with average litter size around six. The geometric progression that followed — common ratio about 7 per year for the first decade — produced an estimated 10 billion rabbits across Australia by 1900.
The continent's grasslands collapsed; native fauna was driven to extinction in many regions. The eventual control measure (the myxoma virus in 1950) reduced the population by 99% — a GP working in reverse, with common ratio about 0.01. See the Australian Government NMRC summary of European rabbit history for the historical record.
The Mathematicians Who Shaped Sequences and Series
Archimedes of Syracuse (c. 287–212 BCE) computed the infinite sum of $1/4 + 1/16 + 1/64 + \cdots = 1/3$ — possibly the first rigorous infinite-GP sum in history. He used it to find the area under a parabolic segment.
Leonardo Fibonacci (c. 1170–1250, Italy) worked with both geometric and arithmetic progressions in Liber Abaci (1202) and laid the groundwork for European understanding of these sequences.
Carl Friedrich Gauss (1777–1855, Germany) generalised series-summation techniques as a child — famously summing 1 to 100 in seconds — and later applied analogous methods to infinite GPs in his work on hypergeometric series.
The story worth keeping. Archimedes solved the infinite-GP sum problem more than 2,000 years before calculus was invented. He did it by what we now call the method of exhaustion — a logical proof that the sum could not exceed $1/3$ and could not be less than $1/3$, so it had to equal $1/3$. The same proof, in modernised notation, is in your Class 11 textbook.
Conclusion
A geometric progression is a sequence where each term is the previous term multiplied by a fixed common ratio $r$.
The $n$th term is $a_n = a \cdot r^{n-1}$.
The finite sum is $S_n = a(1 - r^n)/(1 - r)$; the infinite sum is $a/(1 - r)$ when $|r| < 1$.
The single most common mistake is using the arithmetic-sum formula on a GP — the answers will be wildly off.
GPs model compound interest, population growth, radioactive decay, and music intervals — anything that multiplies by a fixed factor each step.
Try It Yourself — Three Problems
Find the 10th term of the GP $3, 6, 12, 24, \ldots$.
Find the sum of the first 6 terms of $4, 12, 36, 108, \ldots$.
A ball bounces to $1/2$ its previous height each time. Dropped from 16 m, what total vertical distance does it travel before stopping?
Want a live Bhanzu trainer to walk through more GP problems? Book a free demo class — online globally.
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