Formulas · Mathematics

Sequences and progressions

5 formulas, each with worked examples. Revision sheet · Practise these formulas

Class 9

The n-th term of an arithmetic progression (a linear pattern)

a_{n} = a + (n - 1) d

What each letter means
$a_{n}$	the n-th term: the number in position n
$a$	the first term
$n$	the position: 1 for the first term, 2 for the second, and so on
$d$	the common difference: the same jump from each term to the next (negative when the terms go down)

In a list where every jump is the same, the number in any position is the first number plus that jump, taken once for every step after the first.

Why it works

Going from the 1st term to the n-th term takes n − 1 jumps, and each jump adds d. So the n-th term is the first term plus (n − 1) lots of d.

When to use it

A list goes up (or down) by the same amount each time, such as tiles in a growing pattern, savings that rise by a fixed amount, or seats in rows that each have 2 more, and you want a term far along without writing them all.

How to use it

Find the first term a.
Find the jump d: any term minus the one before it.
Put a, d and the position n into a + (n − 1)d.

To remember: To reach the n-th term, you jump n − 1 times.

Other forms

Rearranged $a_{n} = n d + (a - d)$ The same rule sorted by n: n times the jump, plus a fixed number. This is the form Chapter 2 finds for linear patterns (for 1, 3, 5, … it is 2n − 1).
Rearranged $n = \frac{a _{n} - a}{d} + 1$ Which position a given number is in. If n is not a whole number, that number is not in the list.
Rearranged $d = \frac{a _{n} - a}{n - 1}$ The jump, from the first term and any later term.
Special case $a_{1} = a$ When n = 1: the first term needs no jump.

Worked examples

Story

Riya saves ₹50 in the first week and each week she saves ₹10 more than the week before (₹50, ₹60, ₹70, …). How much does she save in week 15?

$a_{15} = 50 + (15 - 1) \times 10$
$= 50 + 140 = 190$
In week 15 she saves ₹190.

Answer: 190

Picture

A growing L shape of tiles has 1 tile in stage 1, 3 in stage 2 and 5 in stage 3; each stage adds one tile to each arm. How many tiles are in stage 100?

Each stage adds 2 tiles, so d = 2, and the first stage has a = 1.
$a_{100} = 1 + (100 - 1) \times 2$
$= 1 + 198 = 199$

Answer: 199

Direct

Find the 20th term of 3, 7, 11, 15, …

$d = 7 - 3 = 4$
$a_{20} = 3 + (20 - 1) \times 4$
$= 3 + 19 \times 4 = 3 + 76 = 79$

Answer: 79

Reverse

Which term of 5, 9, 13, … is 101?

Try it first, then show the working

$5 + (n - 1) \times 4 = 101$
$(n - 1) \times 4 = 96$
$n - 1 = 24$
$n = 25$

Answer: 25

Exam

The 3rd term of an arithmetic progression is 12 and its 10th term is 40. Find the first term and the common difference.

Try it first, then show the working

The 3rd term is a + 2d = 12 and the 10th term is a + 9d = 40.
Subtracting the first from the second: 7d = 28, so d = 4.
Then a = 12 − 2 × 4 = 4.
Check: the 10th term is 4 + 9 × 4 = 40.

Answer: first term 4, common difference 4

Common mistake: Using n jumps instead of n − 1 (the first term needs no jump at all), or taking d as the first term minus the second, which flips its sign.

Class 9

Sum of the first n natural numbers (triangular numbers)

S = \frac{n ( n + 1 )}{2}

What each letter means
$S$	the sum 1 + 2 + 3 + … + n, which is also the n-th triangular number
$n$	the last number added (how many numbers are added)

Adding 1 + 2 + … + n gives n times (n + 1), halved. The same numbers count the dots in triangles: 1, 3, 6, 10, 15, …

Why it works

Write the sum forwards and backwards, one under the other. Each column adds to n + 1 (1 + n, 2 + (n − 1), and so on), and there are n columns, so twice the sum is n(n + 1). As a picture, two equal staircases of dots fit together into an n by (n + 1) rectangle.

When to use it

Adding a long run of whole numbers from 1, counting dots in a triangular pattern, handshakes, stacked logs or rows of seats that grow by one.

How to use it

Find n, the last number in the run from 1.
Multiply n by the next number, n + 1.
Halve the product.

To remember: Forwards plus backwards: n pairs, each worth n + 1.

Other forms

Rearranged $n = \frac{- 1 + 1 + 8 S}{2}$ How many numbers from 1 give a certain sum (it must come out whole).
Same formula, another formThe sum from 1 to n is the n-th triangular number: t₁ = 1, t₂ = 3, t₃ = 6, and each is the one before plus n.

Worked examples

Story

Logs are stacked with 1 log in the top row, 2 in the next, and so on, with 20 logs in the bottom row. How many logs are there?

$S = \frac{20 \times 21}{2} = 210$
There are 210 logs.

Answer: 210

Picture

Dots are arranged in a triangle with 1 dot in the first row, 2 in the second, and 8 rows in all. How many dots are there?

Two copies of the triangle fit together into an 8 by 9 rectangle of 72 dots.
$S = \frac{8 \times 9}{2} = 36$

Answer: 36

Direct

Find 1 + 2 + 3 + … + 10.

$S = \frac{10 \times 11}{2} = 55$

Answer: 55

Reverse

Starting from 1, how many consecutive natural numbers must be added to make 78?

Try it first, then show the working

$n (n + 1) = 156$
n = 12 works, since 12 × 13 = 156 (the other solution, −13, is not a natural number).

Answer: 12

Exam

Find the sum of all the natural numbers from 51 to 100.

Try it first, then show the working

Sum from 51 to 100 = (sum from 1 to 100) − (sum from 1 to 50).
$\frac{100 \times 101}{2} - \frac{50 \times 51}{2} = 5050 - 1275 = 3775$

Answer: 3775

Common mistake: Using n(n − 1) ÷ 2 (that is the sum up to n − 1), or forgetting to halve.

Class 9

The n-th term of a geometric progression

t_{n} = a r^{n - 1}

What each letter means
$t_{n}$	the n-th term: the number in position n
$a$	the first term
$r$	the common ratio: the fixed number each term is multiplied by to get the next
$n$	the position: 1 for the first term, 2 for the second, and so on

In a list where each term is the one before multiplied by the same number r, the n-th term is the first term multiplied by r, n − 1 times.

Why it works

Going from the 1st term to the n-th takes n − 1 steps, and each step multiplies by r. Multiplying by r, n − 1 times, is multiplying by r to the power n − 1.

When to use it

Anything that grows or shrinks by the same factor each step: doubling, a bouncing ball, repeated halving, population growth, patterns like the Sierpiński triangle.

How to use it

Find the first term a.
Find r: any term divided by the one before it.
Put a, r and the position n into a × r^(n − 1).

To remember: To reach the n-th term, multiply n − 1 times.

Other forms

Special case $t_{n} = a$ When r = 1: every term is the same.
Same formula, another formThe common ratio is any term divided by the term before it: r = t(n + 1) ÷ t(n).

Worked examples

Story

A ball bounces up to 3/4 of its previous height each time, and its first bounce reaches 18 feet. How high is the 5th bounce?

$t_{5} = 18 \times (\frac{3}{4})^{4} = 18 \times \frac{81}{256} = \frac{729}{128}$
That is about 5.7 feet.

Answer: 729/128

Picture

In the Sierpiński triangle, stage 1 is one shaded triangle, and at each stage every shaded triangle is replaced by 3 smaller ones. How many shaded triangles are there at stage 6?

$t_{6} = 1 \times 3^{5} = 243$

Answer: 243

Direct

Find the 8th term of 3, 6, 12, 24, …

r = 6 ÷ 3 = 2
$t_{8} = 3 \times 2^{7} = 3 \times 128 = 384$

Answer: 384

Reverse

Which term of 2, 6, 18, … is 1458?

Try it first, then show the working

$2 \times 3^{n - 1} = 1458$
$3^{n - 1} = 729 = 3^{6}$
So n − 1 = 6, and n = 7.

Answer: 7

Exam

A ball is dropped from 24 feet and each bounce reaches 3/4 of the height before. After how many bounces does it first stay below 1/6 of the height it was dropped from?

Try it first, then show the working

1/6 of 24 feet is 4 feet. After k bounces the height is 24 × (3/4)ᵏ.
$24 \times (\frac{3}{4})^{6} = \frac{2187}{512}$
That is about 4.27 feet, still above 4; the 7th bounce reaches about 3.20 feet.
So from the 7th bounce on, it stays below 4 feet.

Answer: 7 bounces

Common mistake: Using r to the power n (one factor too many), or adding r each time as in an arithmetic progression.

Class 10

Sum of the first n terms of an arithmetic progression

S = \frac{n}{2} (2 a + (n - 1) d)

What each letter means
$S$	the sum of the first n terms
$n$	how many terms are added
$a$	the first term
$d$	the common difference (negative when the terms go down)

The sum of an arithmetic progression is the number of terms times the average of the first and last terms; written with a and d, that is n/2 × (2a + (n − 1)d).

Why it works

Write the sum forwards and backwards and add the two lines term by term: every pair adds to the same total, a + (the last term) = 2a + (n − 1)d. There are n such pairs, and that is twice the sum.

When to use it

Adding many terms of a list that goes up or down by a fixed amount: savings, rows of seats or plants, stacked logs, instalments; or finding how many terms give a certain total.

How to use it

Find a, d and n.
Work out 2a + (n − 1)d.
Multiply by n and halve.

To remember: Half the number of terms, times first plus last.

Other forms

Same formula, another formSince the last term is l = a + (n − 1)d, the sum is also S = n/2 × (a + l).
Special case $S = \frac{n ( n + 1 )}{2}$ When 1 + 2 + … + n (a = 1, d = 1).

Worked examples

Story

Asha saves ₹100 in the first month and ₹20 more each month than the month before. How much has she saved after 12 months?

$S = \frac{12}{2} (200 + 11 \times 20)$
$= 6 \times 420 = 2520$
She has saved ₹2520.

Answer: 2520

Picture

A spiral is made of 13 half circles one after another, with radii 0.5 cm, 1 cm, 1.5 cm, … . What is its total length?

A half circle of radius r has length πr, so the lengths are 0.5π, 1π, 1.5π, …, 6.5π: an AP.
Their sum is π × 13/2 × (0.5 + 6.5) = π × 45.5.
That is 45.5π cm, about 143 cm with π ≈ 22/7.

Answer: 45.5π cm, about 143 cm

Direct

Find the sum of the first 22 terms of 8, 3, −2, …

d = 3 − 8 = −5.
$S = \frac{22}{2} (16 + 21 \times (- 5))$
$= 11 \times (16 - 105) = 11 \times (- 89) = - 979$

Answer: −979

Reverse

How many terms of 24, 21, 18, … must be taken for the sum to be 78?

Try it first, then show the working

78 = n/2 × (48 + (n − 1)(−3)), which gives 3n² − 51n + 156 = 0, or n² − 17n + 52 = 0.
(n − 4)(n − 13) = 0, so n = 4 or n = 13.
Both work: the terms from the 5th to the 13th add to zero.

Answer: 4 or 13

Exam

The sum of the first 14 terms of an AP is 1050 and its first term is 10. Find the 20th term.

Try it first, then show the working

1050 = 14/2 × (20 + 13d) = 140 + 91d, so 91d = 910 and d = 10.
$a_{20} = 10 + 19 \times 10 = 200$

Answer: 200

Common mistake: Using n instead of n − 1 inside the bracket, or forgetting that d is negative for a decreasing list (8, 3, −2, … has d = −5).

Class 10

Sum of an arithmetic progression from its first and last terms

S = \frac{n}{2} (a + l)

What each letter means
$S$	the sum of all the terms
$n$	how many terms there are
$a$	the first term
$l$	the last term

When you know the first and last terms, the sum is simply the number of terms times their average: n/2 × (first + last).

Why it works

Pair the first term with the last, the second with the second last, and so on: in an AP every pair adds to the same total a + l. There are n/2 pairs.

When to use it

Sums where the last term is easy to see: 1 to 1000, all two-digit odd numbers, rows of plants from 23 down to 5, rungs of a ladder.

How to use it

Count the terms n (use the n-th term formula if needed).
Add the first and last terms.
Multiply by n and halve.

To remember: Number of terms times the average of first and last.

Other forms

Same formula, another formPutting l = a + (n − 1)d gives the other form, S = n/2 × (2a + (n − 1)d).

Worked examples

Story

A flower bed has 23 rose plants in the first row, 21 in the second, 19 in the third, and so on, with 5 plants in the last row. How many plants are there?

Rows: 23 − 2(n − 1) = 5 gives n − 1 = 9, so there are 10 rows.
$S = \frac{10}{2} (23 + 5) = 5 \times 28 = 140$

Answer: 140

Picture

A ladder's rungs are 25 cm apart and shrink evenly from 45 cm at the bottom to 25 cm at the top. The top and bottom rungs are 2.5 m apart. How much wood do the rungs need?

2.5 m = 250 cm, and 250 ÷ 25 = 10 gaps, so there are 11 rungs.
$S = \frac{11}{2} (45 + 25) = \frac{11}{2} \times 70 = 385$
The rungs need 385 cm of wood.

Answer: 385

Direct

Find 1 + 2 + 3 + … + 1000.

$S = \frac{1000}{2} (1 + 1000)$
$= 500 \times 1001 = 500500$

Answer: 500500

Reverse

An AP starts at 5, ends at 45 and its terms add to 400. How many terms does it have?

Try it first, then show the working

$400 = \frac{n}{2} (5 + 45)$
$400 = 25 n$
$n = 16$

Answer: 16

Exam

Find the sum of all the two-digit odd numbers.

Try it first, then show the working

They are 11, 13, …, 99: an AP with a = 11 and d = 2.
From 11 + 2(n − 1) = 99: n − 1 = 44, so n = 45.
$\frac{45}{2} (11 + 99) = \frac{45}{2} \times 110 = 2475$

Answer: 2475

Common mistake: Miscounting the terms: from 11 to 99 in steps of 2 there are 45 odd numbers, not 44 or 89.