Formulas · Mathematics

Sequences and series

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

Class 11

Sum of the first n terms of a GP

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

What each letter means
$S$	the sum of the first n terms
$a$	the first term
$r$	the common ratio (not 1)
$n$	how many terms are added

In a geometric progression each term is the one before times r. The first n terms add up to a(rⁿ − 1)/(r − 1).

Why it works

Write S = a + ar + ar² + … + arⁿ⁻¹ and multiply by r: rS = ar + ar² + … + arⁿ. Subtracting, almost everything cancels: rS − S = arⁿ − a, so S(r − 1) = a(rⁿ − 1).

When to use it

Adding many terms of a doubling or shrinking list: savings that grow by a percentage, a population, a bouncing ball, grains on a chessboard.

How to use it

Find a (the first term), r (divide any term by the one before) and n (how many terms).
Work out rⁿ, take 1 away, multiply by a, and divide by r − 1.
When r is less than 1, the form a(1 − rⁿ)/(1 − r) keeps the numbers positive.

To remember: First term times (ratio to the n, minus one), over (ratio minus one).

Other forms

Same formula, another form $S = \frac{a ( 1 - r ^{n} )}{1 - r}$ The same, for r less than 1 (top and bottom both multiplied by −1).
Special case $S = na$ When r = 1 (every term is a).

Worked examples

Story

Asha saves ₹100 in the first month and doubles her saving every month. How much has she saved in 10 months?

Here a = 100, r = 2 and n = 10.
$\frac{100 ( 2 ^{10} - 1 )}{2 - 1} = 102300$

Answer: 102300

Picture

A square of side 1 m has a smaller square drawn inside by joining the midpoints of its sides, and so on. Each square has half the area of the one before. Find the total area of the first 5 squares.

Areas: 1, ½, ¼, …, so a = 1 and r = ½.
$\frac{1 ( 1 - ( \frac{1}{2} ) ^{5} )}{1 - \frac{1}{2}} = \frac{31}{16}$

Answer: 1.9375

Direct

Find the sum of the first 6 terms of the GP 3, 6, 12, …

Here a = 3, r = 2 and n = 6.
$\frac{3 ( 2 ^{6} - 1 )}{2 - 1} = 189$

Answer: 189

Reverse

The first 4 terms of a GP with ratio 3 add up to 200. Find the first term.

Try it first, then show the working

200 = a(3⁴ − 1)/(3 − 1) = 40a.
$\frac{200}{40} = 5$

Answer: 5

Exam

How many terms of the GP 3, 3/2, 3/4, … are needed to give the sum 3069/512?

Try it first, then show the working

a = 3, r = ½: 3(1 − (½)ⁿ)/(½) = 6(1 − (½)ⁿ) = 3069/512.
1 − (½)ⁿ = 3069/3072, so (½)ⁿ = 3/3072 = 1/1024.
$2^{10} = 1024$

Answer: 10 terms

Common mistake: Using n − 1 as the power: the sum of n terms has rⁿ, although the n-th term has rⁿ⁻¹.

Class 11

Sum of an infinite GP

S = \frac{a}{1 - r}

What each letter means
$S$	the sum of all the terms, for ever
$a$	the first term
$r$	the common ratio, between −1 and 1

When the ratio is between −1 and 1, the terms shrink towards 0 and the sum of all of them, for ever, settles at a/(1 − r).

Why it works

The sum of n terms is a(1 − rⁿ)/(1 − r). When −1 < r < 1, rⁿ gets closer and closer to 0 as n grows, so the sum gets closer and closer to a/(1 − r).

When to use it

Repeating decimals (0.333… = 3/10 + 3/100 + …), a ball that bounces for ever, and any process that keeps adding smaller and smaller parts.

How to use it

Check that r is between −1 and 1; otherwise there is no finite sum.
Divide the first term by 1 − r.

To remember: First over one minus the ratio.

Other forms

Rearranged $a = S (1 - r)$ The first term from the sum and the ratio.

Worked examples

Story

A ball is dropped from 10 m and each bounce reaches ⅗ of the height before. How far does it travel in all, up and down?

Down 10 m, then each bounce goes up and down: 2 × (6 + 3.6 + …).
Bounces: a = 6, r = ⅗: 6/(1 − ⅗) = 15, so 2 × 15 = 30.
$10 + 2 \times \frac{6}{1 - \frac{3}{5}} = 40$

Answer: 40 m

Picture

Squares are drawn inside each other, each with half the area of the one before, starting with 8 cm². What is the total area of all of them, for ever?

Here a = 8 and r = ½.
$\frac{8}{1 - \frac{1}{2}} = 16$

Answer: 16

Direct

Find the sum of 1 + ½ + ¼ + ⅛ + … for ever.

Here a = 1 and r = ½.
$\frac{1}{1 - \frac{1}{2}} = 2$

Answer: 2

Reverse

An infinite GP has ratio ⅓ and sum 9. Find the first term.

Try it first, then show the working

a = S(1 − r)
$9 (1 - \frac{1}{3}) = 6$

Answer: 6

Exam

Write 0.777… (7 repeating) as a fraction.

Try it first, then show the working

0.777… = 7/10 + 7/100 + 7/1000 + …, a GP with a = 7/10 and r = 1/10.
$\frac{\frac{7}{10}}{1 - \frac{1}{10}} = \frac{7}{9}$

Answer: 7/9

Common mistake: Using it when r is 1 or more (the sum grows without end), or dividing by r − 1 instead of 1 − r.

Class 11

Sum of the first n squares

S = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}

What each letter means
$S$	the sum 1² + 2² + … + n²
$n$	how many squares are added

The squares 1, 4, 9, 16, … up to n² add up to n(n + 1)(2n + 1)/6.

Why it works

Add k³ − (k − 1)³ = 3k² − 3k + 1 for k = 1 to n: the left side cancels down to n³, and the right side is 3S − 3n(n + 1)/2 + n. Solving for S gives n(n + 1)(2n + 1)/6.

When to use it

Adding squares quickly, sums like Σk(k + 1), and counting squares on a grid (an n by n board has 1² + 2² + … + n² squares of all sizes).

How to use it

Multiply n, n + 1 and 2n + 1.
Divide by 6 (the product always divides exactly).

To remember: n, the next one, and twice plus one, over six.

Other forms

To remember itCheck with n = 2: 1 + 4 = 5, and 2 × 3 × 5 ÷ 6 = 5.

Worked examples

Story

Oranges are stacked in a square pyramid: 1 on top, then 4, 9, … with 8 layers. How many oranges are there?

The layers hold 1², 2², …, 8² oranges.
$\frac{8 \times 9 \times 17}{6} = 204$

Answer: 204

Picture

How many squares of all sizes are there on an 8 by 8 chessboard?

There are 8² squares of side 1, 7² of side 2, …, 1² of side 8.
$\frac{8 \times 9 \times 17}{6} = 204$

Answer: 204

Direct

Find 1² + 2² + … + 10².

$\frac{10 \times 11 \times 21}{6} = 385$

Answer: 385

Reverse

The sum of the first n squares is 650. Find n.

Try it first, then show the working

n(n + 1)(2n + 1) = 3900; try n = 12: 12 × 13 × 25 = 3900.
$\frac{12 \times 13 \times 25}{6} = 650$

Answer: 12

Exam

Find the sum 1 × 2 + 2 × 3 + … + 10 × 11.

Try it first, then show the working

k(k + 1) = k² + k, so the sum is 385 + 55.
$385 + \frac{10 \times 11}{2} = 440$

Answer: 440

Common mistake: Squaring the sum of 1 to n instead: that gives the sum of cubes, not squares.

Class 11

Sum of the first n cubes

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

What each letter means
$S$	the sum 1³ + 2³ + … + n³
$n$	how many cubes are added

The cubes 1, 8, 27, … up to n³ add up to the square of 1 + 2 + … + n, that is (n(n + 1)/2)².

Why it works

Add k⁴ − (k − 1)⁴ = 4k³ − 6k² + 4k − 1 for k = 1 to n: the left side is n⁴, and using the sums of k² and k on the right leaves 4S = n²(n + 1)². So S = (n(n + 1)/2)².

When to use it

Adding cubes quickly, and sums built from cubes such as Σk²(k + 1).

How to use it

Find 1 + 2 + … + n = n(n + 1)/2.
Square it.

To remember: The sum of the cubes is the square of the sum.

Other forms

To remember itCheck with n = 3: 1 + 8 + 27 = 36, and (1 + 2 + 3)² = 36.

Worked examples

Story

Cubes of side 1, 2, 3, …, 6 cm are made from 1 cm unit cubes. How many unit cubes are used?

A cube of side k uses k³ unit cubes.
$(\frac{6 \times 7}{2})^{2} = 441$

Answer: 441

Picture

Show with n = 4 that 1³ + 2³ + 3³ + 4³ is a perfect square.

$1 + 8 + 27 + 64 = 100$
$(\frac{4 \times 5}{2})^{2} = 100$

Answer: 100

Direct

Find 1³ + 2³ + … + 10³.

The sum 1 + 2 + … + 10 is 55.
$5 5^{2} = 3025$

Answer: 3025

Reverse

The sum of the first n cubes is 2025. Find n.

Try it first, then show the working

√2025 = 45 = n(n + 1)/2, so n(n + 1) = 90 and n = 9.
$(\frac{9 \times 10}{2})^{2} = 2025$

Answer: 9

Exam

Find 11³ + 12³ + … + 20³.

Try it first, then show the working

The first 20 cubes minus the first 10 cubes.
$21 0^{2} - 5 5^{2} = 41075$

Answer: 41075

Common mistake: Forgetting to square: n(n + 1)/2 is only the sum of 1 to n.

Class 11

The geometric mean of two numbers

G = ab

What each letter means
$G$	the geometric mean
$a$	the first positive number
$b$	the second positive number

The geometric mean of two positive numbers is the square root of their product. Put between them, it makes a GP: a, G, b.

Why it works

For a, G, b to be a GP the ratios must match: G/a = b/G. Cross-multiplying gives G² = ab, so G = √(ab).

When to use it

Inserting a term between two numbers of a GP, average growth rates, and comparing with the arithmetic mean (A ≥ G for positive numbers).

How to use it

Multiply the two numbers.
Take the square root.

To remember: Arithmetic adds and halves; geometric multiplies and roots.

Other forms

Rearranged $b = \frac{G ^{2}}{a}$ The other number from the mean and one number.

Worked examples

Story

A savings fund grew by a factor of 1.21 over two years. By what single factor did it grow each year, on average?

The yearly factor f satisfies f × f = 1.21, the geometric mean of 1 and 1.21.
$1.21 = 1.1$

Answer: 1.1, that is 10% a year

Picture

A rectangle is 16 cm by 4 cm. Find the side of a square with the same area.

The square's side is the geometric mean of the sides.
$16 \times 4 = 8$

Answer: 8

Direct

Find the geometric mean of 4 and 9.

$4 \times 9 = 6$

Answer: 6

Reverse

The geometric mean of 5 and a number is 15. Find the number.

Try it first, then show the working

b = G²/a
$\frac{1 5 ^{2}}{5} = 45$

Answer: 45

Exam

Insert a number between 2 and 18 so that the three make a GP, and compare it with their arithmetic mean.

Try it first, then show the working

G = √(2 × 18) = 6, giving 2, 6, 18 (ratio 3).
The arithmetic mean is 10, larger than 6, as A ≥ G always holds.
$2 \times 18 = 6$

Answer: 6 (the arithmetic mean, 10, is larger)

Common mistake: Halving the sum instead: that is the arithmetic mean, (a + b)/2.