Arithmetic Progressions — nth Term, Sum & Word Problems for Class 10

What is an Arithmetic Progression?

Take a look at these sequences: 2, 5, 8, 11, 14… or 100, 95, 90, 85… What’s common between them? Each term differs from the previous one by a fixed amount. That fixed amount is what makes a sequence an Arithmetic Progression (AP).

More precisely: a sequence $a_1, a_2, a_3, \ldots$ is an AP if $a_{n+1} - a_n$ is constant for all $n$ . That constant is called the common difference, denoted $d$ .

Why does this matter? APs model real-world situations everywhere — a taxi that charges ₹10 per km after a fixed base fare, a savings plan where you deposit the same amount monthly, even the way marks are distributed across a scoring topic in CBSE. The moment you see “equal increments” in a word problem, AP is almost certainly the tool.

This chapter carries solid weightage in Class 10 boards — typically 8-10 marks across 2-3 questions in CBSE. ICSE tests it similarly, often with a 6-mark sums question. If you understand the two core formulas well, this is one of the most reliable scoring chapters in the paper.

Key Terms & Definitions

First term ( $a$ or $a_1$ ): The starting value of the sequence. Example: In 3, 7, 11, 15 — the first term is $a = 3$ .

Common difference ( $d$ ): The fixed difference between consecutive terms. Always calculated as: $d = a_2 - a_1 = a_3 - a_2$

A crucial observation: $d$ can be positive (increasing AP), negative (decreasing AP), or zero (all terms equal — still technically an AP).

General term ( $a_n$ ): The $n$ th term of the AP. This is the single most useful formula in the chapter.

Number of terms ( $n$ ): How many terms the AP contains. Finite APs have a last term, often called $l$ .

Arithmetic Mean (AM): If $a$ , $b$ , $c$ are in AP, then $b$ is the AM of $a$ and $c$ , meaning $b = \dfrac{a+c}{2}$ .

Core Formulas

a_n = a + (n-1)d

Where:

$a$ = first term
$d$ = common difference
$n$ = position of the term

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

Equivalent form when last term $l = a_n$ is known:

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

a_n = S_n - S_{n-1}

This identity rescues you in problems where sum is given but individual terms aren’t.

Methods & Step-by-Step Concepts

Finding the nth Term

The formula $a_n = a + (n-1)d$ works by thinking of each term as “start at $a$ , then add $d$ exactly $(n-1)$ times.”

Steps:

Identify $a$ (first term) and $d$ (second minus first)
Substitute into $a_n = a + (n-1)d$
Simplify

Before using the formula, always verify $d$ is constant using at least two pairs: $a_2 - a_1$ and $a_3 - a_2$ . A sequence can fool you if you only check once.

Finding Sum of n Terms

The derivation Gauss famously did as a child: write $S_n$ forwards and backwards, add them. Each pair sums to $2a + (n-1)d$ , and there are $n$ such pairs, giving $S_n = \frac{n}{2}[2a + (n-1)d]$ .

When to use which form:

Use $\frac{n}{2}[2a + (n-1)d]$ when you know $a$ , $d$ , $n$
Use $\frac{n}{2}(a + l)$ when you know the first and last term (common in “find sum of all multiples of 5 between 1 and 200” type questions)

Finding Number of Terms

If $a_n = l$ (last term is given), set $a + (n-1)d = l$ and solve for $n$ .

The most common error: students forget that $n$ must be a positive integer. If you solve and get $n = 7.5$ or $n = -3$ , there’s a calculation error somewhere — re-check.

Solved Examples

Example 1 — Easy (CBSE Level)

Q: Find the 15th term of the AP: 4, 9, 14, 19, …

Solution:

Identify: $a = 4$ , $d = 9 - 4 = 5$

a_{15} = 4 + (15-1) \times 5 = 4 + 70 = 74

Example 2 — Easy (CBSE Level)

Q: How many terms are in the AP: 7, 13, 19, …, 205?

Solution:

$a = 7$ , $d = 6$ , $a_n = 205$

7 + (n-1) \times 6 = 205

(n-1) \times 6 = 198

n - 1 = 33 \implies n = 34

Example 3 — Medium (CBSE Board Exam Pattern)

Q: The 3rd term of an AP is 4 and its 9th term is −8. Find the first term and common difference. Also find the sum of first 15 terms.

Solution:

We have two equations:

a_3 = a + 2d = 4 \quad \text{...(i)}

a_9 = a + 8d = -8 \quad \text{...(ii)}

Subtract (i) from (ii):

6d = -12 \implies d = -2

Substitute back: $a + 2(-2) = 4 \implies a = 8$

Now find $S_{15}$ :

S_{15} = \frac{15}{2}[2(8) + 14(-2)] = \frac{15}{2}[16 - 28] = \frac{15}{2} \times (-12) = -90

When two terms are given, always set up two equations and subtract to eliminate $a$ . This is a guaranteed pattern in CBSE 3-mark questions — appear confident, write both equations clearly before solving.

Example 4 — Medium (NCERT Exercise 5.3)

Q: Find the sum of all odd numbers between 1 and 100.

Solution:

The odd numbers form the AP: 3, 5, 7, …, 99

$a = 3$ , $d = 2$ , $l = 99$

First find $n$ : $99 = 3 + (n-1) \times 2 \implies n = 49$

S_{49} = \frac{49}{2}(3 + 99) = \frac{49}{2} \times 102 = 49 \times 51 = 2499

Example 5 — Hard (CBSE 2023 / JEE Main Level)

Q: The ratio of the sum of $m$ terms to the sum of $n$ terms of an AP is $m^2 : n^2$ . Show that the ratio of the $m$ th to the $n$ th term is $(2m-1):(2n-1)$ .

Solution:

Given: $\dfrac{S_m}{S_n} = \dfrac{m^2}{n^2}$

\frac{S_m}{S_n} = \frac{\frac{m}{2}[2a + (m-1)d]}{\frac{n}{2}[2a + (n-1)d]} = \frac{m^2}{n^2}

$\frac{2a + (m-1)d}{2a + (n-1)d} = \frac{m}{n} \quad \text{...(}\bigstar\text{)}$

Now, $a_m = a + (m-1)d$ and $a_n = a + (n-1)d$ .

Key trick: Replace $m$ with $2m-1$ and $n$ with $2n-1$ in (★):

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

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

\therefore \frac{a_m}{a_n} = \frac{2m-1}{2n-1} \qquad \blacksquare

This exact proof appeared in CBSE Board 2023 (Set 1) as a 3-mark question. The key step — substituting $2m-1$ for $m$ in the sum ratio — is non-obvious but extremely elegant. Once you see the trick, you’ll never forget it.

Exam-Specific Tips

CBSE Class 10

Marking scheme: 1-mark (identify AP, find $d$ ), 2-mark (find $a_n$ or $S_n$ ), 3-mark (two-variable problems, proofs), 4-mark (word problems with complete setup).
Always show the formula first, then substitute. CBSE awards step marks — if your arithmetic goes wrong but the setup is right, you still get 2 out of 3.
Word problems involving “sum of $n$ terms” almost always need $S_n = \frac{n}{2}(a + l)$ form. Identify the AP clearly in your working.

ICSE Class 10

ICSE tends to test APs within a larger “sequence and series” question. Expect a mix: find the AP, find specific terms, then find a sum — all in one 6-mark question.
The AM property ( $b = \frac{a+c}{2}$ ) appears frequently in ICSE inserted means questions.

JEE Main

APs appear in sequences & series questions — often combined with geometric progressions. A common JEE pattern: three numbers in AP are given as $a-d$ , $a$ , $a+d$ (see below). This symmetry is worth memorising.
Sum of AP questions sometimes appear disguised as “find the value of $1 + 3 + 5 + \ldots + (2n-1)$ ” — that sum equals $n^2$ .

JEE Main trick: When three unknowns are in AP, always take them as $a-d$ , $a$ , $a+d$ . Their sum is $3a$ (the $d$ cancels). This converts a 3-variable problem into a 1-variable problem immediately.

Common Mistakes to Avoid

Mistake 1: Using $n$ instead of $(n-1)$ in the formula. The formula is $a + (n-1)d$ , not $a + nd$ . The first term needs zero additions of $d$ . Write the formula on your rough work every time until it’s automatic.

Mistake 2: Getting $d$ wrong from word problems. “Each year, salary increases by ₹500” — students often set $d = 500$ when salary increases, but if the problem asks “how much less is term 5 than term 1?”, they get confused about sign. Always define AP clearly: what is $a_1$ , what is $a_2$ .

Mistake 3: Confusing $S_n$ and $a_n$ . $S_{10}$ means the sum of first 10 terms. $a_{10}$ means the 10th term alone. Use the identity $a_n = S_n - S_{n-1}$ only when sum is given — don’t apply it universally.

Mistake 4: Non-integer $n$ accepted as answer. If you’re finding “how many terms does this AP have” and get a decimal, stop — your formula application has an error. Number of terms is always a positive whole number.

Mistake 5: Using the wrong form of the sum formula. When last term $l$ is given, use $S_n = \frac{n}{2}(a + l)$ . Many students blindly use the longer form and waste time expanding $l = a + (n-1)d$ . Identify which values you have before choosing the formula.

Practice Questions

Q1. Which of the following are APs? Find $d$ for those that are. (a) 2, 4, 8, 16, … (b) −5, −1, 3, 7, … (c) $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots$

(a) $4-2=2$ , $8-4=4$ — differences not equal. Not an AP.

(b) $d = -1-(-5) = 4$ . Check: $3-(-1)=4$ ✓. AP with $d = 4$ .

Q2. Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.

$a_{11} = a + 10d = 38$ …(i) $a_{16} = a + 15d = 73$ …(ii)

Subtract: $5d = 35 \implies d = 7$

From (i): $a = 38 - 70 = -32$

$a_{31} = -32 + 30 \times 7 = -32 + 210 = \mathbf{178}$

Q3. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first 3 terms.

$a_4 + a_8 = (a+3d) + (a+7d) = 2a + 10d = 24 \implies a + 5d = 12$ …(i)

$a_6 + a_{10} = (a+5d) + (a+9d) = 2a + 14d = 44 \implies a + 7d = 22$ …(ii)

Subtract (i) from (ii): $2d = 10 \implies d = 5$

$a = 12 - 25 = -13$

First 3 terms: $\mathbf{-13, -8, -3}$

Q4. A sum of ₹700 is to be used to give 7 cash prizes to students. If each prize is ₹20 less than its preceding prize, find the value of each prize.

Let the prizes form an AP with first term $a$ (largest prize) and $d = -20$ .

$S_7 = 700$

\frac{7}{2}[2a + 6(-20)] = 700

\frac{7}{2}[2a - 120] = 700

2a - 120 = 200 \implies a = 160

Prizes: ₹160, ₹140, ₹120, ₹100, ₹80, ₹60, ₹40

Q5. The sum of first $n$ terms of an AP is $3n^2 + 5n$ . Find the AP and its 25th term.

$S_n = 3n^2 + 5n$

$a_1 = S_1 = 3 + 5 = 8$

$a_2 = S_2 - S_1 = (12+10) - 8 = 14$

$d = 14 - 8 = 6$

AP: $8, 14, 20, 26, \ldots$

$a_{25} = 8 + 24 \times 6 = 8 + 144 = \mathbf{152}$

Shortcut: $a_n = S_n - S_{n-1} = 3n^2 + 5n - [3(n-1)^2 + 5(n-1)] = 6n + 2$ . So $a_{25} = 6(25)+2 = 152$ ✓

Q6. Find the sum of all multiples of 7 lying between 100 and 1000.

First multiple of 7 after 100: $7 \times 15 = 105$

Last multiple of 7 before 1000: $7 \times 142 = 994$

AP: 105, 112, …, 994. Here $a = 105$ , $d = 7$ , $l = 994$ .

Find $n$ : $105 + (n-1) \times 7 = 994 \implies n-1 = 127 \implies n = 128$

S = \frac{128}{2}(105 + 994) = 64 \times 1099 = \mathbf{70336}

Q7. If $m$ times the $m$ th term of an AP equals $n$ times the $n$ th term, show that the $(m+n)$ th term is zero.

$m \cdot a_m = n \cdot a_n$

$m[a + (m-1)d] = n[a + (n-1)d]$

$a(m-n) + d[m(m-1) - n(n-1)] = 0$

$a(m-n) + d[m^2 - m - n^2 + n] = 0$

$a(m-n) + d[(m^2-n^2) - (m-n)] = 0$

$a(m-n) + d(m-n)(m+n-1) = 0$

Since $m \neq n$ , divide by $(m-n)$ :

$a + d(m+n-1) = 0$

But $a_{m+n} = a + (m+n-1)d = 0$ $\qquad \blacksquare$

Q8. The ratio of the sums of first $m$ and first $n$ terms of an AP is $m^2:n^2$ . Find the ratio of the $m$ th and $n$ th terms.

This is the classic proof shown in Solved Example 5 above.

Answer: $a_m : a_n = (2m-1):(2n-1)$

Key step: In $\frac{S_m}{S_n} = \frac{m}{n}$ , replace $m \to 2m-1$ and $n \to 2n-1$ to convert sum ratio into term ratio.

FAQs

What is the difference between a sequence and an AP?

A sequence is any ordered list of numbers. An AP is a special sequence where consecutive terms have a fixed common difference. All APs are sequences, but not all sequences are APs. The sequence 1, 4, 9, 16 (perfect squares) is not an AP since the differences 3, 5, 7 keep changing.

Can the common difference be zero?

Yes. If $d = 0$ , every term equals $a$ . The sequence 5, 5, 5, 5 is a valid AP with $d = 0$ . This case appears in CBSE objective questions to test whether students know the definition precisely.

Can the common difference be negative?

Absolutely. A decreasing AP has negative $d$ . Example: 50, 45, 40, 35… has $d = -5$ . Temperature dropping by 3°C each hour, savings decreasing each month — real applications have negative $d$ all the time.

How do I find the first term when only $S_n$ is given?

Use $a_1 = S_1$ . Substitute $n = 1$ into the expression for $S_n$ . Then find $S_2$ , and get $a_2 = S_2 - S_1$ . Now $d = a_2 - a_1$ .

When should I use $S_n = \frac{n}{2}(a + l)$ versus the other form?

Use $(a + l)$ form when the last term $l$ is explicitly given (e.g., “sum of all even numbers from 2 to 200”). Use $[2a + (n-1)d]$ form when you have $a$ , $d$ , and $n$ but not the last term explicitly.

Is Arithmetic Mean the same as average?

For two numbers, yes — AM of $a$ and $c$ is $\frac{a+c}{2}$ , which is their average. In AP context, if $a$ , $b$ , $c$ are in AP then $b$ is the AM. The CBSE Class 10 syllabus covers this as a property, not a separate formula.

What does “insert $n$ arithmetic means between $a$ and $b$ ” mean?

You’re creating a sequence $a, A_1, A_2, \ldots, A_n, b$ — total $(n+2)$ terms forming an AP. The common difference is $d = \frac{b-a}{n+1}$ . ICSE tests this frequently; CBSE Class 10 rarely does (it appears more in Class 11).

Is the sum formula derived from Gauss’s trick?

Yes, exactly. Write $S_n = a + (a+d) + \ldots + l$ forward, then reverse: $S_n = l + (l-d) + \ldots + a$ . Add the two rows: $2S_n = n(a+l)$ . This is why understanding the derivation helps — if you forget the formula under exam pressure, you can re-derive it in 30 seconds.