Sequence and Series — AFCAT Numerical Ability Formulas and Traps

Arithmetic Progression: Terms, Counting Terms and Sums

The nth term of an AP

nth term of an AP

a_n = a + (n - 1)d

Which term is it, and how many terms are there

Number of terms

n = \dfrac{l - a}{d} + 1

An AP from two of its terms

Difference from two terms

d = \dfrac{a_q - a_p}{q - p}

Sum of the first n terms

Sum of n terms of an AP

S_n = \dfrac n2 (a + l) = \dfrac n2 [2a + (n - 1)d]

From a sum formula to the nth term

nth term from the sum

a_n = S_n - S_{n-1}

Symmetric terms and the middle term

Three terms in AP

a - d,\; a,\; a + d \qquad \text{sum} = 3a

Common traps

n minus 1 steps, not n

The 10th term of

3, 8, 13, \dots

is

3 + 9 \times 5 = 48

. Using

10 \times 5

gives 53, one step too far.

Subtract in order

d

is a later term minus the earlier one. For

20, 17, 14

it is

17 - 20 = -3

. Reading it as 3 sends the progression the wrong way.

Add the 1

From 6 to 102 in steps of 4 there are 24 steps but 25 terms. Dividing the gap by

d

counts the steps, not the terms.

A fraction means no

If solving gives

n - 1 = 17.4

, the number is not in the AP at all. Do not round it to the nearest term.

Count the steps, not the positions

Between the 4th and 10th terms there are

10 - 4 = 6

steps. Dividing the gap by 7, or by 10, gives the wrong

d

.

Two roots, one answer

3n^2 + n - 310 = 0

has roots 10 and

-\tfrac{31}{3}

. A number of terms is a positive whole number, so only 10 counts.

2a inside the bracket

The bracket in

\dfrac n2 [2a + (n - 1)d]

holds twice the first term. Writing

a

there undercounts every pair, and the answer lands on a distractor.

S of n is not the nth term

Putting

n = 6

into the sum formula gives the sum of six terms, not the sixth term. Subtract

S_5

from it.

Centre on the middle term

With

a, a + d, a + 2d

the sum is

3a + 3d

and both unknowns stay. Writing

a - d, a, a + d

makes the sum give

a

at once.

Both signs of d

d^2 = 4

gives

d = 2

or

d = -2

. Both give the same three numbers in opposite order, so neither is lost.

Geometric Progression: Terms, Sums and the Sum to Infinity

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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

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

Common traps

r to the power n minus 1

The 6th term of

3, 6, 12, \dots

is

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.

The power plus one

If

\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.

Two square roots

r^2 = 25

gives

r = 5

or

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.

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

.

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.

Harmonic Progression, and the Three Means

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Terms of an HP through the reciprocals

nth term of an HP

h_n = \dfrac{1}{a + (n - 1)d}

The harmonic mean of two numbers

Harmonic mean

H = \dfrac{2ab}{a + b}

AM, GM and HM together, and inserted means

Order of the three means

A \ge G \ge H, \qquad G^2 = AH

AP, GP or HP: the middle-term test

Progression	Middle term b	Test to check
AP	Arithmetic mean $\dfrac{a + c}{2}$	$2b = a + c$
GP	Geometric mean $\sqrt{ac}$	$b^2 = ac$
HP	Harmonic mean $\dfrac{2ac}{a + c}$	$\dfrac 2b = \dfrac 1a + \dfrac 1c$ The HP test is the AP test on the reciprocals.

Common traps

Flip back at the end

In

\tfrac 12, \tfrac 16, \tfrac 1{10}, \dots

the 8th reciprocal is 30, so the 8th term is

\tfrac 1{30}

. Answering 30 forgets the last step.

The terms themselves do not step evenly

In the HP

\tfrac 17, \tfrac 1{14}, \tfrac 1{21}

the gaps are

\tfrac 1{14}

and

\tfrac 1{42}

. Only the reciprocals

7, 14, 21

go up by the same amount.

Not the average of the reciprocals

For 6 and 10 the reciprocals average to

\tfrac 2{15}

, so the harmonic mean is

\tfrac{15}{2} = 7.5

. Stopping at

\tfrac 2{15}

answers a different question.

The order matters

The test is on the middle term in the order given.

2, 3, 6

is an HP, but

6, 2, 3

is not: the harmonic mean of 6 and 3 is 4, not 2.

Equal numbers pass every test

5, 5, 5

is an AP, a GP and an HP at once. So 'both an AP and a GP' means the three numbers are equal.

The order is fixed

For two different positive numbers, AM is greater than GM, which is greater than HM. An option that puts HM above GM is wrong without any working.

G is not the average of A and H

G^2 = AH

. For 3 and 12,

\tfrac{7.5 + 4.8}{2} = 6.15

, but the geometric mean is 6.

Special Series and the Standard Sums

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Series built from known pieces

Split and sum

\sum_{k=1}^{n} (u_k + v_k) = \sum_{k=1}^{n} u_k + \sum_{k=1}^{n} v_k

The standard sums

Sum	Result	Example
$1 + 2 + 3 + \dots + n$	$\dfrac{n(n + 1)}{2}$	$1 + 2 + \dots + 40 = 820$
$1 + 3 + 5 + \dots + (2n - 1)$	$n^2$	The first 15 odd numbers add up to $225$
$2 + 4 + 6 + \dots + 2n$	$n(n + 1)$	$2 + 4 + \dots + 40 = 420$
$1^2 + 2^2 + \dots + n^2$	$\dfrac{n(n + 1)(2n + 1)}{6}$	$1^2 + 2^2 + \dots + 10^2 = 385$
$1^3 + 2^3 + \dots + n^3$	$\left[\dfrac{n(n + 1)}{2}\right]^2$	$1^3 + 2^3 + \dots + 10^3 = 3025$ The sum of the cubes is the square of the sum of the numbers.

Common traps

Count the odd numbers

The odd numbers from 1 to 79 are 40 numbers, not 79 and not 39, so their sum is

40^2 = 1600

.

Even numbers are not n squared

2 + 4 + \dots + 2n = n(n + 1)

, one

n

more than the odd-number sum

n^2

. The first 10 even numbers add up to 110, not 100.

Pairs need an even count

Pairing

1^2 - 2^2 + 3^2 - \dots

works cleanly only for an even number of terms. With an odd number the last square is left over and has to be added back.

A GP of n terms, not an infinite one

In

n - (0.1 + 0.01 + \dots)

the bracket stops after

n

terms, so it is

\tfrac 19 (1 - 10^{-n})

. Using the sum to infinity,

\tfrac 19

, is only right when

n

is huge.

Sequence and Series formulas

Arithmetic Progression: Terms, Counting Terms and Sums

The nth term of an AP

Which term is it, and how many terms are there

An AP from two of its terms

Sum of the first n terms

From a sum formula to the nth term

Symmetric terms and the middle term

Common traps

Geometric Progression: Terms, Sums and the Sum to Infinity

The nth term of a GP

Which term is it, and how many terms

A GP from two of its terms, and the middle term

Sum of the first n terms

Common traps

Harmonic Progression, and the Three Means

Terms of an HP through the reciprocals

The harmonic mean of two numbers

AM, GM and HM together, and inserted means

AP, GP or HP: the middle-term test

Common traps

Special Series and the Standard Sums

Series built from known pieces

The standard sums

Common traps

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