CBSE Class 10 Maths — Arithmetic Progressions

Chapter Overview & Weightage

Arithmetic Progressions (AP) is one of the highest-scoring chapters in CBSE Class 10 Maths. The concepts are few and well-defined, the formulas are compact, and the question types are predictable. Students who master the two main formulas — nth term and sum — and practise word problems can score full marks consistently.

AP carries 6–8 marks in CBSE Class 10 board exams. Typical distribution: 1 MCQ (1 mark) + 1 short answer finding nth term or sum (2 marks) + 1 long answer word problem (3–4 marks). The word problem is the key scoring question — learn to identify AP in context.

Year	Marks	Question Types
2024	7	MCQ + nth term + word problem (sum of salary)
2023	8	MCQ + AR-type + find first term/difference + sum
2022	6	Short answer: verify AP, find common difference + nth term
2021	7	Identify AP, sum formula application

Key Concepts You Must Know

1. Arithmetic Progression (AP): A sequence where the difference between consecutive terms is constant.

General form: $a, a+d, a+2d, a+3d, \ldots$
$a$ = first term, $d$ = common difference

2. Common Difference: $d = a_2 - a_1 = a_3 - a_2 = \ldots$ (must be constant)

$d > 0$ : increasing AP
$d < 0$ : decreasing AP
$d = 0$ : constant sequence (special case)

3. General (nth) Term: $a_n = a + (n-1)d$

4. Sum of First n Terms:

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

where $l = a_n$ is the last term. The second form is faster when first and last terms are known.

5. Relationship between nth term and sum: $a_n = S_n - S_{n-1}$

Important Formulas

a_n = a + (n-1)d

Use when: finding any specific term, finding which term has a given value.

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

Use when: finding total of a sequence, problems about savings, distances, etc.

a_n = S_n - S_{n-1}

Use when: sum formula is given and you need individual terms.

Solved Previous Year Questions

PYQ 1 — CBSE 2024 (3 marks)

The sum of first 7 terms of an AP is 49, and the sum of first 17 terms is 289. Find the sum of first $n$ terms.

Solution:

$S_7 = \frac{7}{2}[2a + 6d] = 49 \implies 2a + 6d = 14 \implies a + 3d = 7$ …(i)

$S_{17} = \frac{17}{2}[2a + 16d] = 289 \implies 2a + 16d = 34 \implies a + 8d = 17$ …(ii)

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

From (i): $a = 7 - 6 = 1$

S_n = \frac{n}{2}[2(1) + (n-1)(2)] = \frac{n}{2}[2 + 2n - 2] = \frac{n}{2} \times 2n = n^2

$S_n = n^2$

Verify: $S_7 = 49 = 7^2$ ✓, $S_{17} = 289 = 17^2$ ✓

PYQ 2 — CBSE 2023 (2 marks)

How many terms of the AP 3, 5, 7, 9, … are needed for their sum to equal 120?

Solution: $a = 3$ , $d = 2$

$S_n = 120 \implies \frac{n}{2}[2(3) + (n-1)(2)] = 120$

$\frac{n}{2}[6 + 2n - 2] = 120 \implies \frac{n}{2}[2n + 4] = 120$

$n(n + 2) = 120 \implies n^2 + 2n - 120 = 0$

$(n + 12)(n - 10) = 0 \implies n = 10$ (rejecting $n = -12$ )

10 terms are needed.

PYQ 3 — Word Problem (4 marks)

A sum of Rs 1000 is invested at 8% per annum simple interest. Find the interest at the end of 1st, 2nd, 3rd, … year. Is it an AP? Find the interest at the end of 30 years.

Solution:

Simple Interest for year 1 = $\frac{1000 \times 8 \times 1}{100} = 80$

Interest for year 2 = 80; for year 3 = 80; …

Each year adds Rs 80. This is an AP with $a = 80$ , $d = 0$ (constant sequence).

Interest at end of 30 years (total SI for 30 years) = $30 \times 80 = 2400$

Or using AP sum: $S_{30} = \frac{30}{2}[2(80) + 29(0)] = 15 \times 160 = 2400$

Difficulty Distribution

Difficulty	Topics	Approximate %
Easy	Identify AP, find common difference, write next terms	30%
Medium	Find nth term, sum formula, find which term equals value	45%
Hard	Word problems (salary, savings, distances), two-variable problems	25%

Expert Strategy

Identify the AP pattern in word problems. The most common exam trap is recognising an AP in context:

Salary increasing by fixed amount each year → AP
Ball bouncing to fixed fraction of height → NOT AP (GP)
Seats increasing by fixed number each row → AP
Distance run each day increasing by 1 km → AP

For two-variable problems ( $a$ and $d$ ): Always set up two equations using the given information about specific terms or sums, then solve simultaneously.

When three numbers are in AP, write them as $a-d$ , $a$ , $a+d$ (not $a$ , $a+d$ , $a+2d$ ). This halves your algebra — when you add three AP terms, the middle one doubles: $(a-d) + a + (a+d) = 3a$ . This trick saves significant time in CBSE problems where the sum of three AP terms is given.

Sum of all odd (or even) terms: Use the sum formula with $n$ = number of such terms, and identify the appropriate AP for odd/even positions.

Common Traps

Trap 1: Using $a_n = a + nd$ instead of $a + (n-1)d$ . The nth term formula starts at $n=1$ giving $a_1 = a + (1-1)d = a$ (correct). Using $a + nd$ gives $a_1 = a + d$ (wrong — that’s $a_2$ ). Always double-check by substituting $n=1$ .

Trap 2: Rejecting valid negative solutions for number of terms. When solving $n^2 + 2n - 120 = 0$ , you get $n = 10$ or $n = -12$ . Always reject negative $n$ — number of terms must be a positive integer. But don’t reject non-integer solutions silently — if $n$ comes out to 10.5, recheck your setup.

Trap 3: Confusing $S_n$ and $a_n$ . Some problems give $S_n$ as a quadratic and ask for $a_n$ . Use $a_n = S_n - S_{n-1}$ . Do NOT differentiate $S_n$ — that’s for calculus, not AP. Verify: $a_1 = S_1$ (not $S_1 - S_0$ , unless you define $S_0 = 0$ ).

Trap 4: Arithmetic error in verifying AP. To check if a sequence is AP, the difference must be constant — not just equal between consecutive pair. For 3, 7, 12, 18: differences are 4, 5, 6 — not constant, so this is NOT an AP. Always check at least two consecutive differences.