AFCAT Numerical Ability · Sequence and Series

Geometric Progression: Terms, Sums and the Sum to Infinity

A sequence that is multiplied by the same number at every step: find any term, which term a number is, the progression from two of its terms, the sum of n terms and the sum to infinity.

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Why this matters

The questions mirror the AP page with multiplying in place of adding, and the same slip costs marks: the ratio is used one time fewer than the term number. The new idea is the sum to infinity, which exists only when the ratio lies between minus 1 and 1. The HARD questions run the formulas backwards to find the number of terms, the first term or the ratio.

Concept 1 of 5: The nth term of a GP

A geometric progression (GP) multiplies by the same number, the common ratio

r

, at every step. From the first term to the nth there are

n - 1

steps, so

r

is used

n - 1

times.

Definition

The common ratio is any term divided by the one before it: $r = \dfrac{a_2}{a_1}$ . It can be negative or a fraction.
The nth term is $a_n = ar^{n-1}$ .
A negative ratio makes the signs alternate: odd powers of $r$ are negative, even powers positive.
Given a term and the ratio, the first term is $a = \dfrac{a_n}{r^{n-1}}$ .

nth term of a GP

a_n = a r^{n-1}

Worked example

Find the 6th term of the GP

3, 6, 12, \dots

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 1 · Sequence and Series · Geometric Progression

a = 5

r = -2

, find the

4^{th}

term of the GP.

r to the power n minus 1

The 6th term of

3, 6, 12, \dots

3 \times 2^5 = 96

. Using

2^6

gives 192, the 7th term.

Bracket a negative ratio

With

r = -3

(-3)^4 = 81

but

-3^4 = -81

. Drop the bracket and every even-numbered term comes out with the wrong sign.

Drill 7 more on the nth term of a gp

Concept 2 of 5: Which term is it, and how many terms

Divide the number by the first term and ask which power of

r

is left. If

\dfrac xa = r^k

, the number is term

k + 1

, because the first term already carries

r^0

Definition

$x = ar^{n-1}$ means $r^{n-1} = \dfrac xa$ . Write that as a power of $r$ and read off $n - 1$ .
If $\dfrac xa$ is not a whole power of $r$ , $x$ is not a term.
Number of terms from $a$ to the last term $l$ : $r^{n-1} = \dfrac la$ , then add 1 to the power.
Powers worth knowing: $2^{10} = 1024$ , $3^5 = 243$ , $4^5 = 1024$ , $5^4 = 625$ .

Position of a term

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

Worked example

Which term of the GP

5, 10, 20, \dots

is 640?

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionHard

Example 2 · Sequence and Series · Geometric Progression

Which term of the GP

2, 6, 18, 54, ...

1458

The power plus one

\dfrac xa = 2^7

, the number is the 8th term, not the 7th. The power counts the steps, and the first term is one term before any step.

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Concept 3 of 5: A GP from two of its terms, and the middle term

Dividing one term by another cancels

a

and leaves a power of

r

: the 7th term over the 4th is

r^3

. Multiplying two terms that sit equally far from a middle term gives the middle term squared.

Definition

$\dfrac{a_q}{a_p} = r^{q-p}$ . Find $r$ , then the first term $a = \dfrac{a_p}{r^{p-1}}$ .
An even power has two real roots: $r^2 = 9$ gives $r = 3$ or $r = -3$ . Keep both unless the question rules one out.
$a, b, c$ are in GP exactly when $b^2 = ac$ : the middle one is the geometric mean $\sqrt{ac}$ of the other two.
More generally $a_{m-k} \times a_{m+k} = (a_m)^2$ for any $k$ .
When the terms are expressions in an unknown, set $b^2 = ac$ and solve. Reject a value that makes a term zero.

Ratio from two terms

\dfrac{a_q}{a_p} = r^{\,q-p}

Worked example

The 2nd term of a GP is 15 and the 5th term is 405. Find the common ratio and the first term.

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionHard

Example 3 · Sequence and Series · Geometric Progression

The

3^{rd}

term of a GP is

16

and

6^{th}

term is

128

. Find the first term.

Two square roots

r^2 = 25

gives

r = 5

r = -5

. Writing only 5 loses an answer when the options list both signs, or list them together as

\pm 5

No term can be zero

When

b^2 = ac

gives two values of

x

, test each in the terms. A value that makes a term zero, such as

x = -2

when one term is

x + 2

, is not allowed: use the other root.

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Concept 4 of 5: Sum of the first n terms

Multiply the sum by

r

and subtract the original: every term cancels except the first and the one just past the last. That leaves

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

Definition

$S_n = \dfrac{a(r^n - 1)}{r - 1}$ . For $r < 1$ the same thing reads more easily as $\dfrac{a(1 - r^n)}{1 - r}$ .
If $r = 1$ , every term is $a$ and $S_n = na$ .
Given the sum and asked for $n$ : solve for $r^n$ first, then read off the power.
The sum uses $r^n$ , while the nth term uses $r^{n-1}$ .

Sum of n terms of a GP

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

Worked example

Find the sum of the first 6 terms of the GP

5, 15, 45, \dots

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionHard

Example 4 · Sequence and Series · Geometric Progression

How many terms of the GP

1, 2, 4, 8, ...

have sum equal to

255

r to the n, not n minus 1

The sum of 6 terms uses

r^6

; the 6th term uses

r^5

. Mixing them up is the commonest slip on this page.

Subtract 1 before multiplying

With

a = 5

r = 2

and

n = 4

, the sum is

5(2^4 - 1) = 75

, not

5 \times 2^4 = 80

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Concept 5 of 5: Sum to infinity

When the ratio lies between minus 1 and 1, the terms shrink towards zero and the running total settles at a fixed value. Start at 3 and take a third each time:

3 + 1 + \tfrac 13 + \dots

never passes 4.5.

Definition

$S_\infty = \dfrac{a}{1 - r}$ , valid only when $-1 < r < 1$ .
For $r \ge 1$ or $r \le -1$ the terms do not shrink and there is no sum to infinity.
Given $S_\infty$ and $a$ : $1 - r = \dfrac{a}{S_\infty}$ , so $r = 1 - \dfrac{a}{S_\infty}$ .
A negative ratio works too: $r = -\tfrac 13$ gives $S_\infty = \dfrac{a}{4/3} = \tfrac{3a}{4}$ .
The sum of $n$ terms in terms of the sum to infinity: $S_n = S_\infty(1 - r^n)$ .

Sum to infinity

S_\infty = \dfrac{a}{1 - r}, \quad -1 < r < 1

Worked example

Find the sum to infinity of

9 + 3 + 1 + \tfrac 13 + \dots

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionHard

Example 5 · Sequence and Series · Geometric Progression

If the sum of an infinite GP is

12

and first term is

6

, find the common ratio.

Only when the terms shrink

2 + 4 + 8 + \dots

has no sum to infinity. Putting

r = 2

into

\dfrac{a}{1 - r}

gives

-2

, which means nothing.

1 minus r, not r minus 1

The denominator is

1 - r

. For

9 + 3 + 1 + \dots

\dfrac{9}{r - 1}

gives

-13.5

, a negative sum of positive terms.

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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)

The nth term of a GP
nth term of a GP
$a_n = a r^{n-1}$
Which term is it, and how many terms
Position of a term
$r^{n-1} = \dfrac{a_n}{a}$
A GP from two of its terms, and the middle term
Ratio from two terms
$\dfrac{a_q}{a_p} = r^{\,q-p}$
Sum of the first n terms
Sum of n terms of a GP
$S_n = \dfrac{a(r^n - 1)}{r - 1}, \quad r \ne 1$
Sum to infinity
Sum to infinity
$S_\infty = \dfrac{a}{1 - r}, \quad -1 < r < 1$

Watch out for (9)

r to the power n minus 1
→ The nth term of a GP
Bracket a negative ratio
→ The nth term of a GP
The power plus one
→ Which term is it, and how many terms
Two square roots
→ A GP from two of its terms, and the middle term
No term can be zero
→ A GP from two of its terms, and the middle term
r to the n, not n minus 1
→ Sum of the first n terms
Subtract 1 before multiplying
→ Sum of the first n terms
Only when the terms shrink
→ Sum to infinity
1 minus r, not r minus 1
→ Sum to infinity

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