Sequences & Progressions

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Arithmetic Progressions

⏱ 5 min read🧩 6 question types🎯 13 practice Q

The idea in one minute

An arithmetic progression (AP) is a list of numbers that grows or falls by the same fixed amount, the common difference $d$ . Its $n$ -th term is $a + (n-1)d$ , and its $n$ -term sum is $\dfrac{n}{2}$ times (first + last). Row-of-seats, monthly savings and wage-step questions are APs in disguise.

What an arithmetic progression is

An arithmetic progression (AP) is a list of numbers with one simple rule. Every term comes from the previous term by adding the same fixed number.

That fixed number is the common difference, written $d$ . The first term is $a$ .

Seats in a hall: row 1 has 12 seats, row 2 has 15, row 3 has 18. Here $a = 12$ and $d = 3$ .

Rule: Before using any formula, check $d$ . Subtract any two neighbouring terms — the answer must be the same everywhere.

A falling list like 40, 36, 32 is also an AP. Its $d$ is $-4$ . A difference can be zero, positive or negative.

The nth term

The $n$ -th term is the first term plus $n - 1$ jumps of size $d$ .

a_n = a + (n - 1)d

Why $n - 1$ ? From term 1 to term 10 there are 9 jumps, not 10.

The 10th term of 7, 11, 15, ... is $7 + 9 \times 4 = 43$ .

The last of $n$ terms is $a + (n-1)d$ — the same formula counted from the other end.

Tip: In word problems, "the 10th row" means $n = 10$ . The 10th row is $a + 9d$ , never $a + 10d$ .

The sum of n terms

Add the first and the last term. An AP makes every such pair equal: $a_1 + a_n = a_2 + a_{n-1}$ , and so on.

There are $n$ such pairs, shared between two terms at a time. Half of $n$ pairs, each worth first + last:

S_n = \frac{n}{2}\left[2a + (n-1)d\right] \qquad S_n = \frac{n}{2}(\text{first} + \text{last})

Sum of 20 terms of 5, 9, 13, ...: $\dfrac{20}{2}[10 + 19 \times 4] = 10 \times 86 = 860$ .

The middle term trick

For an odd number of terms, the middle term equals the average.

\text{middle} = \frac{S_n}{n}

The sum of 15 terms is 375, so the 8th (middle) term is $\dfrac{375}{15} = 25$ .

Three consecutive terms are best named $a - d$ , $a$ , $a + d$ . Their sum is $3a$ , so $a$ is the sum divided by 3.

The sum of three numbers in AP is 27 and their product is 504. Middle $= 9$ , so $9(81 - d^2) = 504$ , giving $d = 5$ : the numbers are 4, 9, 14.

Example: Terms equally far from the ends add to the same value: $a_2 + a_{15} = a_1 + a_{16}$ .

Inserting arithmetic means

Numbers placed between two numbers so that the whole list becomes an AP are arithmetic means (AMs).

With $k$ means between $a$ and $b$ , the $k$ means create $k + 1$ equal jumps:

d = \frac{b - a}{k + 1}

Insert 4 means between 8 and 23: $d = \dfrac{15}{5} = 3$ , so the means are 11, 14, 17, 20.

Watch: Divide by $k + 1$ , not by $k$ . Four means between two numbers make five gaps.

Word problems: rows, savings, wages

Translate the story into $a$ , $d$ and $n$ first.

First month or first row gives $a$ .
The fixed increase every step gives $d$ .
A total asks for $S_n$ ; the value of step $n$ is $a + (n-1)d$ .

A man saves Rs 200 in the first month and adds Rs 25 every month. Twelve months: $S_{12} = \dfrac{12}{2}[400 + 11 \times 25] = 6 \times 675 = 4050$ .

Seats: 15 rows, first row 15 chairs, each row 2 more. Last row $= 15 + 14 \times 2 = 43$ ; total $= \dfrac{15}{2}(15 + 43) = 435$ .

Tip: Say what each symbol means in the story before touching the formulas.

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common2 practice Q

nth term of an AP

How to spot it:

A list with a constant gap is given, and a specific term (5th, 10th, last) is asked.

a_n = a + (n-1)d

Method

Read $a$ and $d$ from the list.
Count the jumps: $n - 1$ .
Add $d$ $(n-1)$ times to $a$ .

Why it works:

Each step of the list adds exactly one $d$ , and reaching term $n$ takes $n - 1$ steps.

Try this

Find the 10th term of the AP 7, 11, 15, 19, ...

Show solution

$a = 7$ , $d = 4$ .
$a_{10} = 7 + 9 \times 4 = 43$ .

Answer

Type 2very common3 practice Q

Sum of n terms of an AP

How to spot it:

'Find the sum of the first 20 terms' — a total of a constant-gap list is asked.

S_n = \frac{n}{2}\left[2a + (n-1)d\right]

Method

Read $a$ , $d$ and $n$ .
Compute $2a + (n-1)d$ .
Multiply by $n/2$ .

Why it works:

Equal end-pairs let the whole sum collapse to (average term) × (number of terms).

Try this

Find the sum of the first 20 terms of the AP 5, 9, 13, ...

Show solution

$a = 5$ , $d = 4$ , $n = 20$ .
$S = \dfrac{20}{2}[10 + 76] = 10 \times 86$ .

Answer

860

Type 3common2 practice Q

Which term of the AP is this value?

How to spot it:

'Which term of 3, 8, 13, ... is 78?' — a value is given and its position $n$ is asked.

Method

Write $a + (n-1)d =$ target.
Move $a$ across and divide by $d$ .
Add 1 to get $n$ .

Why it works:

The target value fixes the number of $d$ -jumps after the first term.

Try this

Which term of the AP 3, 8, 13, 18, ... is 78?

Show solution

$3 + (n-1)5 = 78$ .
$(n-1) = 15$ , so $n = 16$ .

Answer

16th term

Type 4common2 practice Q

Middle term: sum, average and three-term problems

How to spot it:

The sum of an odd number of AP terms is given and the middle term is asked — or three AP numbers hide behind a sum and a product.

\text{middle} = \frac{S_n}{n}

Method

Divide the sum by the count for the middle term.
For three terms, write $a-d$ , $a$ , $a+d$ .
Use the product (or second condition) to find $d$ .

Why it works:

In an AP the average sits exactly at the middle, and three consecutive terms are symmetric about it.

Try this

The sum of 15 terms of an AP is 375. Find the 8th term.

Show solution

8th is the middle of 15 terms.
$375 \div 15 = 25$ .

Answer

Type 5occasional2 practice Q

Insert arithmetic means between a and b

How to spot it:

'Insert 4 arithmetic means between 8 and 23' — the means, their gap, or one of them is asked.

d = \frac{b - a}{k + 1}

Method

Count the gaps: $k + 1$ for $k$ means.
Find the gap $d = (b-a)/(k+1)$ .
Add $d$ repeatedly from $a$ .

Why it works:

The $k$ inserted numbers split the distance from $a$ to $b$ into $k + 1$ equal jumps.

Try this

Insert 4 arithmetic means between 8 and 23.

Show solution

$d = \dfrac{23 - 8}{5} = 3$ .
Means: 11, 14, 17, 20.

Answer

11, 14, 17, 20

Type 6very common2 practice Q

AP word problems: rows, seats, savings, wages

How to spot it:

Seats grow by a fixed number per row, savings by a fixed amount per month, wages by a fixed step — a total or a later value is asked.

S_n = \frac{n}{2}\left[2a + (n-1)d\right]

Method

Name $a$ (first value), $d$ (fixed step), $n$ (count).
Use $a + (n-1)d$ for one term.
Use $S_n$ for the total.

Why it works:

A story with a constant per-step change is exactly an AP wearing words.

Try this

A hall has 18 rows of chairs; the first row has 15 chairs and each row has 3 more than the previous row. How many chairs are there in all?

Show solution

$a = 15$ , $d = 3$ , $n = 18$ .
$S = \dfrac{18}{2}[30 + 17 \times 3] = 9 \times 81$ .

Answer

729

Formula sheet

nth term

a_n = a + (n-1)d

$a$ is the first term, $d$ the common difference.

Sum of n terms

S_n = \frac{n}{2}\left[2a + (n-1)d\right]

Works for any AP.

Sum from first and last

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

$l$ is the last term.

Middle term

\text{middle} = \frac{S_n}{n}

Only when the count $n$ is odd.

Gap when inserting k means

d = \frac{b - a}{k + 1}

$k$ means create $k + 1$ gaps.

Shortcuts that save time

⚡ The middle term is the average

For an odd count of AP terms, sum $=$ middle $\times$ count. One division gives the middle term, and one multiplication gives the sum.

Example

The sum of 9 terms of an AP is 135. Its middle term is:

Show solution

Middle $= 135 \div 9$ .
$= 15$ .

Answer

⚡ Equidistant terms add to the same value

$a_2 + a_{15}$ , $a_1 + a_{16}$ — pairs placed equally far from the ends share one sum. A sum of all $n$ terms then needs no $a$ or $d$ at all.

Example

In an AP of 16 terms, the sum of the 2nd and 15th terms is 40. Find the sum of all 16 terms.

Show solution

$a_2 + a_{15} = a_1 + a_{16} = 40$ .
$S_{16} = \dfrac{16}{2} \times 40$ .
$= 320$ .

Answer

320

⚡ Name three terms a-d, a, a+d

Three consecutive AP terms collapse to one unknown $a$ plus $d$ . Sum gives $a$ at once; product or another condition then gives $d$ .

Example

The sum of three numbers in AP is 27 and their product is 504. Find the numbers.

Show solution

Middle $= 27 \div 3 = 9$ .
$9(81 - d^2) = 504$ , so $d^2 = 25$ .
$d = 5$ : the numbers are 4, 9, 14.

Answer

4, 9, 14

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Using $a + nd$ for the nth term — the 1st term must come out at $n = 1$ , so the term is $a + (n-1)d$ .

Mistake 02

Writing the sum as $\dfrac{n}{2} \times$ last term — the rule is $\dfrac{n}{2} \times$ (first term + last term).

Mistake 03

Using $(b - a)/k$ as the gap when inserting $k$ arithmetic means — the $k$ means create $k + 1$ steps, so the gap is $(b-a)/(k+1)$ .

Mistake 04

In word problems, taking the 10th row as $a + 10d$ — row 1 is $a$ itself, so row 10 is $a + 9d$ .

Mistake 05

Forgetting that an AP may have a negative common difference — terms then fall, and the sum can shrink as $n$ grows.

Mistake 06

Looking for a middle term when the count is even — no single middle term exists, so the $S_n/n$ shortcut does not apply.

Quick revision

Read this the night before the exam.

nth term: $a + (n-1)d$ — $n-1$ jumps, not $n$ .
Sum: $\dfrac{n}{2}[2a + (n-1)d]$ or $\dfrac{n}{2}(\text{first} + \text{last})$ .
Middle term of an odd-count AP $= S_n \div n$ .
Terms equidistant from the ends add to the same value.
Inserting $k$ means: gap $= \dfrac{b-a}{k+1}$ .
Three AP terms: $a-d$ , $a$ , $a+d$ ; their sum is $3a$ .

Practice: 13 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 13 questions

Suggested time 10 min · wrong answers go to your mistake notebook automatically.

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