Geometric progressions

A bacteria culture doubling every hour, a rumour spreading through a hostel, the value of an investment compounding year on year — all three grow the same way: each new value is a fixed multiple of the one before it. That multiplying pattern is a geometric progression (G.P.), the natural counterpart to the arithmetic progression, where terms grow by repeated addition instead of repeated multiplication. The difference matters enormously: add a constant amount every step and growth is a straight line; multiply by a constant ratio every step and growth explodes (or collapses toward zero) exponentially. This article builds the three tools ISC Class 11 expects for a G.P. — the general term, the sum of $n$ terms, and the sum to infinity — plus the geometric mean and the classic geometric-mean-insertion problem, all grounded in one running numerical example so you see exactly how the formulas are used, not just what they say.

The core insight

A geometric progression is a sequence in which every term is the previous term multiplied by a fixed common ratio $r$ , so the $n$ -th term is $T_n = ar^{n-1}$ and the sum of its first $n$ terms is $S_n = \dfrac{a(r^n-1)}{r-1}$ , with a finite sum to infinity $S_\infty = \dfrac{a}{1-r}$ only when $|r| < 1$ .

The core idea

A sequence $a_1, a_2, a_3, \ldots$ is a geometric progression if the ratio of every term to the one immediately before it is the same fixed nonzero number, the common ratio $r$ :

r = \frac{a_{n+1}}{a_n} \quad \text{for every } n,

where: $a_n$ is the $n$ -th term and $r$ can be positive (steady growth or decay), negative (terms alternate in sign), or a fraction (terms shrink toward zero). If the first term is $a$ , the sequence reads $a, ar, ar^2, ar^3, \ldots$ , so the exponent on $r$ is always one less than the term number. The general ( $n$ -th) term is

T_n = ar^{n-1}.

Summing the first $n$ terms gives (proved below, for $r \ne 1$ )

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

where: the two forms are algebraically identical — use whichever keeps the denominator positive. When $r=1$ every term equals $a$ , so the formula's zero denominator is avoided by using $S_n = na$ directly instead. If $|r| < 1$ , repeated multiplication by $r$ drives $r^n \to 0$ as $n \to \infty$ , so an infinite G.P. can still have a finite total, the sum to infinity:

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

This is why $1 + \tfrac12 + \tfrac14 + \tfrac18 + \cdots$ (here $a=1$ , $r=\tfrac12$ ) settles at the finite value $S_\infty = \dfrac{1}{1-\tfrac12} = 2$ , rather than growing without bound the way $1+2+4+8+\cdots$ does. If three positive numbers $a, b, c$ are in G.P., the middle term is the geometric mean (G.M.) of the outer two:

b = \sqrt{ac}.

To insert $n$ geometric means $G_1, \ldots, G_n$ between positive numbers $a$ and $b$ so that $a, G_1, \ldots, G_n, b$ forms one G.P., treat $b$ as the $(n+2)$ -th term of that progression:

r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}}, \qquad G_k = ar^k \text{ for } k = 1, \ldots, n.

A.M.–G.M. relation: for any two positive numbers $a$ and $c$ , the arithmetic mean is never less than the geometric mean,

\frac{a+c}{2} \ge \sqrt{ac},

with equality only when $a = c$ — a fact that follows from $(\sqrt a - \sqrt c)^2 \ge 0$ .

Figure

Bar heights for a geometric progression starting at one and doubling five times compared against a straight line for an arithmetic progression with the same start and step

Derivation AThe sum of the first n terms of a geometric progression

The goal is to find a closed formula for $S_n = a + ar + ar^2 + \cdots + ar^{n-1}$ , the sum of the first $n$ terms of a G.P. with first term $a$ and common ratio $r \ne 1$ , using nothing beyond writing the series twice and subtracting — no calculus, just algebra any Class 11 student already owns.

Step 1. Write the sum out in full: $S_n = a + ar + ar^2 + \cdots + ar^{n-2} + ar^{n-1}$ . Every term after the first is $r$ times the one before it, so multiplying the whole equation by $r$ shifts every term one place to the right: $rS_n = ar + ar^2 + ar^3 + \cdots + ar^{n-1} + ar^n$ .

Step 2. Subtract the second line from the first. Almost everything cancels because $rS_n$ contains the same terms $ar, ar^2, \ldots, ar^{n-1}$ that $S_n$ does, just shifted: $S_n - rS_n = a - ar^n$ , since every middle term appears once in $S_n$ and once in $rS_n$ and cancels, leaving only the first term of $S_n$ (which is $a$ ) and the last term of $rS_n$ (which is $ar^n$ , with a minus sign).

Step 3. Factor both sides. The left side is $S_n(1-r)$ , so $S_n(1-r) = a(1-r^n)$ . Since $r \ne 1$ , divide both sides by $(1-r)$ to isolate $S_n$ : $S_n = \dfrac{a(1-r^n)}{1-r}$ , which is identical to $\dfrac{a(r^n-1)}{r-1}$ after multiplying numerator and denominator by $-1$ .

Step 4. Sanity-check on a tiny case. For $a=1$ , $r=2$ , $n=4$ : direct addition gives $1+2+4+8=15$ . The formula gives $S_4 = \dfrac{1(2^4-1)}{2-1} = \dfrac{15}{1} = 15$ . Matches.

Worked example

The 3rd term of a G.P. is 20 and the 6th term is 160. Find the first term, the common ratio, and the sum of the first 6 terms.

Step 1. Write both given terms using $T_n = ar^{n-1}$ : $T_3 = ar^2 = 20$ and $T_6 = ar^5 = 160$ .

Step 2. Divide the second equation by the first to eliminate $a$ : $\dfrac{ar^5}{ar^2} = \dfrac{160}{20}$ , so $r^3 = 8$ , giving $r = 2$ .

Step 3. Substitute $r=2$ back into $ar^2=20$ : $a(2)^2 = 20 \Rightarrow 4a = 20 \Rightarrow a = 5$ . Check: $T_6 = 5(2)^5 = 5 \times 32 = 160$ , which matches the given data.

Step 4. Apply the sum formula with $a=5$ , $r=2$ , $n=6$ : $S_6 = \dfrac{a(r^6-1)}{r-1} = \dfrac{5(2^6-1)}{2-1} = \dfrac{5(64-1)}{1} = 5 \times 63 = 315$ .

The first term is $a=5$ , the common ratio is $r=2$ , and $S_6 = 315$ .

Competitive stretch

Take the same idea one level further with a G.P. that has $|r|<1$ : for $8, -4, 2, -1, \ldots$ , $a=8$ and $r = \dfrac{-4}{8} = -\dfrac12$ . Since $|r|=\tfrac12<1$ , the sum to infinity exists: $S_\infty = \dfrac{a}{1-r} = \dfrac{8}{1-(-\tfrac12)} = \dfrac{8}{\tfrac32} = \dfrac{16}{3}$ . And inserting $2$ geometric means between $2$ and $16$ (so $16$ is the 4th term, $n+1=3$ ) gives $r = (16/2)^{1/3} = 8^{1/3} = 2$ , so the means are $G_1 = 2(2)=4$ and $G_2 = 2(4)=8$ , producing $2,4,8,16$ — each term double the last.

Common traps

Using the A.P. mean formula for a G.P. The geometric mean is $\sqrt{ac}$ , a square root of a product, not $\tfrac{a+c}{2}$ , an average of a sum — these get swapped under exam pressure.
Applying $S_n = \dfrac{a(r^n-1)}{r-1}$ when $r=1$ . This divides by zero; a constant sequence with $r=1$ needs $S_n = na$ instead.
Using $S_\infty = \dfrac{a}{1-r}$ when $|r|\ge 1$ . The series only converges when $|r|<1$ ; for $|r|\ge 1$ the terms never shrink and there is no finite sum.
Off-by-one when inserting geometric means. Inserting $n$ means between $a$ and $b$ makes $b$ the $(n+2)$ -th term, so the ratio uses exponent $\tfrac{1}{n+1}$ , not $\tfrac1n$ .

Ask the tutor: Want to practice setting up two G.P. equations from given terms, or work through another sum-to-infinity problem step by step? Ask the AI tutor to walk through one with you.

How is a geometric progression different from an arithmetic progression in practice?

An A.P. adds the same fixed number every step, so it grows or shrinks in a straight line — the difference between consecutive terms, $d$ , is constant. A G.P. multiplies by the same fixed ratio $r$ every step, so it grows (or decays) exponentially — the ratio of consecutive terms is constant, not the difference. This is why a G.P. can shoot up very fast for $r>1$ or shrink rapidly toward zero for $0<r<1$ , patterns an A.P. can never produce.

Why does the sum to infinity formula only work when the common ratio is a fraction less than 1 in size?

$S_\infty = a/(1-r)$ comes from taking $S_n = a(1-r^n)/(1-r)$ and letting $n \to \infty$ . The only part that changes with $n$ is $r^n$ : if $|r|<1$ , repeated multiplication shrinks $r^n$ toward $0$ , so the sum settles at a fixed finite value. If $|r|\ge 1$ , $r^n$ either stays at size $1$ forever or grows without bound, so the partial sums never settle — there is nothing finite to converge to, and writing down $S_\infty$ in that case is meaningless.

When I insert geometric means between two numbers, why do I use n+1 as the exponent and not just n?

Because inserting $n$ means between $a$ and $b$ creates a longer G.P.: $a$ is the 1st term, the $n$ inserted means are terms $2$ through $n+1$ , and $b$ itself becomes the $(n+2)$ -th term. Using $T_k = ar^{k-1}$ with $k=n+2$ gives $b = ar^{n+1}$ , so solving for $r$ needs the $(n+1)$ -th root, $r = (b/a)^{1/(n+1)}$ . Forgetting that $b$ is one term further along than " $n$ means" suggests is exactly the off-by-one trap this formula guards against.

Geometric progressions

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads