NDA Maths · Sequence & Series

Geometric Progressions — the constant-ratio engine

A list where each term is the one before it times a fixed ratio — so terms grow (or shrink) by multiplication, and an unending GP can still add up to a finite total.

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Why this matters

Nineteen PYQs across 2017–2026, mostly EASY–MODERATE. NDA tests the nth term, the finite sum, and above all the infinite sum (when the ratio lies strictly between minus one and one) — repeating decimals, continued fractions, and infinite products all reduce to it. Two more recurring shapes: GP-preserving operations and the product-of-terms symmetry. Five concepts cover the lot.

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Concept 1 of 6: nth term, geometric mean, and the three-term condition

In a GP every step multiplies by the same number — the common ratio

r

. So the nth term is the first term times

r

raised to

(n-1)

. Three numbers are in GP when the middle is the geometric mean of the outer two: its square equals their product.

Definition

A geometric progression has first term $a$ and common ratio $r = \dfrac{a_{k+1}}{a_k}$ (constant, $r \ne 0$ ). Then:

nth term: $a_n = a\,r^{\,n-1}$ .
Geometric mean of $a$ and $b$ : $\text{GM} = \sqrt{ab}$ .
Three-term condition: $a, b, c$ are in GP $\iff b^2 = ac$ .

nth term and three-term condition

a_n = a\,r^{\,n-1}, \qquad b^2 = ac \ \text{(for } a,b,c \text{ in GP)}

$a$ first term
$r$ common ratio

Worked example

Find the 6th term of the GP

3, 6, 12, 24, \ldots

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2019 · Q1Easy

Example 1 · Sequence & Series · Geometric Progressions

What is the

n^{th}

term of the sequence

25, -125, 625, -3125, \ldots

The GP nth term is $a\,r^{n-1}$ , not $a\,r^{n}$

The first term has the ratio applied zero times, so position

n

carries

r^{n-1}

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

. Using

a\,r^{n}

overshoots by one factor of

r

. For

3, 6, 12, \ldots

the 5th term is

3\cdot 2^{4} = 48

, not

3\cdot 2^{5} = 96

(that is the 6th term).

Drill 2 more on nth term, geometric mean, and the three-term condition

Concept 2 of 6: Sum of a finite GP

Adding up a GP has a one-line formula because almost everything cancels when you subtract

r

times the sum from the sum itself. The same formula handles repeating-digit sums (

0.3 + 0.33 + 0.333 + \cdots

) once you factor each term into a GP.

Definition

For a GP with first term $a$ , ratio $r \ne 1$ , and $n$ terms:

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

Use the first form when

r > 1

, the second when

r < 1

(both are equal). If

r = 1

the GP is constant and

S_n = na

Sum of n terms

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

Worked example

Find the sum of the first 6 terms of the GP

2, 6, 18, \ldots

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2017 · Q8Moderate

Example 2 · Sequence & Series · Geometric Progressions

What is the sum of the series

0.3 + 0.33 + 0.333 + \ldots

n

terms?

Repeating-digit sums hide a GP

For

0.3 + 0.33 + 0.333 + \cdots

, write each term as

\tfrac{3}{9}(1 - 10^{-k})

. The sum splits into a constant part (an AP-like count of

\tfrac13

) and a true GP

\sum 10^{-k}

. Don't try to treat the original list as a GP directly — it isn't one.

Drill 3 more on sum of a finite gp

Concept 3 of 6: Sum of an infinite GP

If the ratio is between

-1

and

1

, the terms shrink toward zero fast enough that the whole unending sum settles on a finite number. That single formula cracks repeating decimals and infinite products (take logs first for products). A periodic continued fraction or nested radical is NOT a GP — those refer to themselves and are solved by a self-referential equation; see the next concept.

Definition

For $|r| < 1$ , the infinite GP converges:

S_\infty = \frac{a}{1 - r}.

|r| \ge 1

the sum does not converge (no finite value). Many problems reverse this: given

S_\infty

and one term, solve for

a

and

r

Sum to infinity

S_\infty = \frac{a}{1 - r}\quad (|r| < 1)

Worked example

Find the sum to infinity of

4 + 2 + 1 + \tfrac12 + \cdots

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Sep 2018 · Q7Easy

Example 3 · Sequence & Series · Geometric Progressions

The sum of the series

3 - 1 + \frac{1}{3} - \frac{1}{9} + \cdots

is equal to

The convergence condition is not optional

S_\infty = \tfrac{a}{1-r}

is only valid for

|r| < 1

. If a problem's ratio has

|r| \ge 1

, the sum genuinely has no finite value — there is no number to find. Always check

|r|

before reaching for the formula.

It is $\dfrac{a}{1-r}$ , watch the sign in the denominator

The infinite sum is

S_\infty = \dfrac{a}{1-r}

— first term over

(1 - r)

. Flipping it to

\dfrac{a}{r-1}

negates the answer. For

4 + 2 + 1 + \cdots

with

r = \tfrac12

S_\infty = \dfrac{4}{1 - \tfrac12} = 8

; the wrong

\dfrac{4}{\tfrac12 - 1} = -8

is negative even though every term is positive.

Drill 5 more on sum of an infinite gp

Concept 4 of 6: Periodic continued fractions & nested radicals

A repeating continued fraction or nested radical contains a copy of ITSELF. The trick is not a GP sum: set the whole expression equal to

x

, notice the part inside is the same whole

x

again, replace it, and solve the (usually quadratic) equation that results.

Definition

For a periodic continued fraction $x = a + \dfrac{1}{a + \dfrac{1}{a + \cdots}}$ , the tail equals the whole, so $x = a + \dfrac{1}{x} \Rightarrow x^2 - ax - 1 = 0$ . For a nested radical $x = \sqrt{a + \sqrt{a + \cdots}}$ , squaring gives $x^2 = a + x$ . In both cases take the positive root (the expression is positive).

Self-referential equations

x = a + \tfrac{1}{x}\ \Rightarrow\ x^2 - ax - 1 = 0, \qquad x = \sqrt{a + x}\ \Rightarrow\ x^2 - x - a = 0

Worked example

Evaluate the golden continued fraction

x = 1 + \dfrac{1}{1 + \dfrac{1}{1 + \cdots}}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Sep 2019 · Q22Moderate

Example 4 · Sequence & Series · Geometric Progressions

What is the value of

2 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2 + \cdots \infty}}}

It is NOT an infinite GP

Reaching for

\tfrac{a}{1-r}

here is wrong — there is no common ratio. The structure is self-referential: set it to

x

, substitute the inner copy, solve the quadratic, keep the positive root.

Concept 5 of 6: What preserves a GP

Operations that act the same way on every term keep a GP a GP. Multiply or divide each term by a constant, raise each to the same power, or take reciprocals — all still GPs (with a new ratio). This is the GP mirror of the AP-preserving rules, and it sets up the log-bridge trick in the next subtopic.

Definition

If $a, b, c$ are in GP (so $b^2 = ac$ ), then so are:

$ka,\ kb,\ kc$ and $\tfrac{a}{k},\ \tfrac{b}{k},\ \tfrac{c}{k}$ (ratio $r$ unchanged);
$a^2,\ b^2,\ c^2$ (ratio $r^2$ ) and $\sqrt{a},\ \sqrt{b},\ \sqrt{c}$ (ratio $\sqrt{r}$ );
$\tfrac1a,\ \tfrac1b,\ \tfrac1c$ (ratio $\tfrac1r$ ).

Adding a constant to each term, however, does NOT preserve a GP.

Worked example

a, b, c

are in GP, prove that

\tfrac1a, \tfrac1b, \tfrac1c

are in GP.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2022 · Q24Easy

Example 5 · Sequence & Series · Geometric Progressions

a, b, c

are in GP where

a>0, b>0, c>0

, then which of the following are correct? 1.

a^2, b^2, c^2

are in GP 2.

\frac{1}{a}, \frac{1}{b}, \frac{1}{c}

are in GP 3.

\sqrt{a}, \sqrt{b}, \sqrt{c}

are in GP Select the correct answer using the code given below:

Drill 1 more on what preserves a gp

Concept 6 of 6: Product of terms and the middle-term trick

Pair up terms of a GP from the two ends: each such pair multiplies to the same value (the square of the middle term). So the product of an odd number of terms is just the middle term raised to that count — no need to know

a

r

individually.

Definition

In a GP, terms equidistant from the ends have a constant product: $a_k \cdot a_{n+1-k} = a_1 \cdot a_n$ . Consequently the product of the first $2m-1$ terms equals the middle term raised to the power $2m-1$ : if the middle term is $M$ , the product is $M^{2m-1}$ . This is why "the kth term is given" is often enough to find a product.

GP product symmetry

a_k \cdot a_{n+1-k} = a_1 \cdot a_n \qquad \prod_{i=1}^{2m-1} a_i = M^{2m-1}

Worked example

The 4th term of a GP is 2. Find the product of its first 7 terms.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2021 · Q53Easy

Example 6 · Sequence & Series · Geometric Progressions

The third term of a GP is 3. What is the product of its first five terms?

Drill 3 more on product of terms and the middle-term trick

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (5)

nth term, geometric mean, and the three-term condition
nth term and three-term condition
$a_n = a\,r^{\,n-1}, \qquad b^2 = ac \ \text{(for } a,b,c \text{ in GP)}$
Sum of a finite GP
Sum of n terms
$S_n = \frac{a\,(r^n - 1)}{r - 1}\quad (r \ne 1)$
Sum of an infinite GP
Sum to infinity
$S_\infty = \frac{a}{1 - r}\quad (|r| < 1)$
Periodic continued fractions & nested radicals
Self-referential equations
$x = a + \tfrac{1}{x}\ \Rightarrow\ x^2 - ax - 1 = 0, \qquad x = \sqrt{a + x}\ \Rightarrow\ x^2 - x - a = 0$
Product of terms and the middle-term trick
GP product symmetry
$a_k \cdot a_{n+1-k} = a_1 \cdot a_n \qquad \prod_{i=1}^{2m-1} a_i = M^{2m-1}$

Watch out for (5)

The GP nth term is $a\,r^{n-1}$ , not $a\,r^{n}$
→ nth term, geometric mean, and the three-term condition
Repeating-digit sums hide a GP
→ Sum of a finite GP
The convergence condition is not optional
→ Sum of an infinite GP
It is $\dfrac{a}{1-r}$ , watch the sign in the denominator
→ Sum of an infinite GP
It is NOT an infinite GP
→ Periodic continued fractions & nested radicals

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