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Practice Set · Class 10

Class 10 Maths Arithmetic Progressions: 35 Important Questions by Question Type

35 original problems, every question type covered, every answer independently verified. Attempt each one before revealing the solution.

By Genelis Team 19 min read Updated 2026-08-27T09:00:00+05:30

Arithmetic Progressions questions repeat across a small, identifiable set of patterns once you've seen the full landscape mapped out. This is 35 original problems, 5 for every distinct type, each with a complete solution you can follow line by line.

How to use this page: Attempt each problem fully on paper before tapping "Reveal Solution" — check your working, not just your final number. Every calculation on this page was computed and independently verified before publishing.

Quick formula reference used throughout:

What it finds Formula
nth term an = a + (n−1)d
Sum of n terms Sn = n/2 [2a + (n−1)d]
mth term from the end l − (m−1)d, where l is the last term

Every Question Type Covered

1

Finding the nth term

2

Number of terms / which term equals a value

3

Sum of n terms

4

Finding an AP from two given terms

5

mth term from the end

6

Verifying an AP from its general term

7

Word problems

Type 1 · Finding the nth Term

1.1

Direct Term

Find the 15th term of the AP: 3, 7, 11, 15, ...

Reveal Solution

Identify a and d

a = 3, d = 4

a₁₅ = 3 + (15−1)×4 = 3 + 56
✓ 15th term = 59
1.2

Negative Common Difference

Find the 22nd term of the AP: 8, 3, −2, −7, ...

Reveal Solution

Identify a and d

a = 8, d = −5

a₂₂ = 8 + (22−1)×(−5) = 8 − 105
✓ 22nd term = −97
1.3

Direct Term

If a = 5 and d = 6, find the 10th term.

Reveal Solution
a₁₀ = 5 + (10−1)×6 = 5 + 54
✓ 10th term = 59
1.4

Solving for Term Number

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

Reveal Solution

Set up the equation

a = 3, d = 5. Solve 3 + (n−1)×5 = 78

(n−1)×5 = 75 → n−1 = 15
✓ n = 16 (the 16th term is 78)
1.5

Direct Term

If a = 10 and d = 4, find the 30th term.

Reveal Solution
a₃₀ = 10 + (30−1)×4 = 10 + 116
✓ 30th term = 126

Type 2 · Number of Terms

2.1

Given Last Term

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

Reveal Solution

a = 7, d = 6, last term = 205

Solve 7 + (n−1)×6 = 205

(n−1)×6 = 198 → n−1 = 33
✓ n = 34 terms
2.2

Counting a Set

How many two-digit numbers are divisible by 3?

Reveal Solution

The sequence is 12, 15, 18, ..., 99

a = 12, d = 3, last term = 99

12 + (n−1)×3 = 99 → n−1 = 29
✓ n = 30 two-digit numbers
2.3

Counting a Set

How many multiples of 4 lie between 10 and 250?

Reveal Solution

The sequence is 12, 16, 20, ..., 248

a = 12, d = 4, last term = 248

12 + (n−1)×4 = 248 → n−1 = 59
✓ n = 60 multiples
2.4

Solving for Term Number

In an AP with a = 5 and d = 3, find n such that the nth term is 101.

Reveal Solution
5 + (n−1)×3 = 101 → (n−1)×3 = 96 → n−1 = 32
✓ n = 33
⚠️ If solving this kind of equation gives a non-integer n, it means the target value simply isn't a term of that AP at all — double-check the target number and the AP's terms before assuming a calculation error.
2.5

Two Valid Answers

How many terms of the AP 63, 60, 57, ... are needed to give a sum of 693?

Reveal Solution

a = 63, d = −3. Set up the sum equation

n/2 [2(63) + (n−1)(−3)] = 693

Simplifying gives a quadratic in n, with solutions n = 21 and n = 22

Check why both work

The 22nd term = 63 + 21×(−3) = 0. Adding a term equal to zero doesn't change the sum — so S₂₁ and S₂₂ are both exactly 693.

✓ n = 21 or n = 22 (both genuinely correct)
⚠️ This isn't a mistake in the algebra — whenever a sum-based equation for an AP gives two positive integer solutions for n, check whether one of the terms between them equals zero. If it does, both values are legitimately correct.

Type 3 · Sum of n Terms

3.1

Direct Sum

Find the sum of the first 20 terms of the AP: 2, 7, 12, 17, ...

Reveal Solution

a = 2, d = 5, n = 20

S₂₀ = 20/2 × [2(2) + (20−1)(5)] = 10 × [4+95]
✓ S₂₀ = 990
3.2

Negative Terms

Find the sum of the first 15 terms of the AP: 8, 5, 2, −1, ...

Reveal Solution

a = 8, d = −3, n = 15

S₁₅ = 15/2 × [2(8) + (15−1)(−3)] = 7.5 × [16−42]
✓ S₁₅ = −195
3.3

Multiples

Find the sum of the first 40 positive multiples of 6.

Reveal Solution

a = 6, d = 6, n = 40

S₄₀ = 40/2 × [2(6) + (40−1)(6)] = 20 × [12+234]
✓ S₄₀ = 4920
3.4

Sum of a Range

Find the sum of all natural numbers from 1 and 100 that are divisible by 4.

Reveal Solution

Sequence: 4, 8, ..., 100. a=4, d=4, last term=100

Find n first: 4+(n−1)4=100 → n=25

S₂₅ = 25/2 × (4+100) = 12.5 × 104
✓ Sum = 1300
3.5

Direct Sum

If a = 7 and d = 4, find the sum of the first 25 terms.

Reveal Solution
S₂₅ = 25/2 × [2(7) + (25−1)(4)] = 12.5 × [14+96]
✓ S₂₅ = 1375

Type 4 · Finding an AP from Two Given Terms

4.1

Two Conditions

The 3rd term of an AP is 16 and the 7th term is 32. Find the AP.

Reveal Solution

Set up two equations

a+2d=16 ... (i), a+6d=32 ... (ii)

Subtracting (i) from (ii): 4d=16 → d=4. Then a=16−2(4)=8
✓ a = 8, d = 4 → AP: 8, 12, 16, 20, ...
4.2

Two Conditions

The 4th term of an AP is 0 and the 9th term is 20. Find the AP.

Reveal Solution

Set up two equations

a+3d=0 ... (i), a+8d=20 ... (ii)

Subtracting: 5d=20 → d=4. Then a=−3(4)=−12
✓ a = −12, d = 4 → AP: −12, −8, −4, 0, 4, ...
4.3

Two Conditions

The 6th term of an AP is 12 and the 10th term is 20. Find the AP.

Reveal Solution

Set up two equations

a+5d=12 ... (i), a+9d=20 ... (ii)

Subtracting: 4d=8 → d=2. Then a=12−5(2)=2
✓ a = 2, d = 2 → AP: 2, 4, 6, 8, ...
4.4

Two Conditions

The 5th term of an AP is 19 and the 12th term is 47. Find the AP.

Reveal Solution

Set up two equations

a+4d=19 ... (i), a+11d=47 ... (ii)

Subtracting: 7d=28 → d=4. Then a=19−4(4)=3
✓ a = 3, d = 4 → AP: 3, 7, 11, 15, ...
4.5

Two Conditions

The 8th term of an AP is 39 and the 15th term is 74. Find the AP.

Reveal Solution

Set up two equations

a+7d=39 ... (i), a+14d=74 ... (ii)

Subtracting: 7d=35 → d=5. Then a=39−7(5)=4
✓ a = 4, d = 5 → AP: 4, 9, 14, 19, ...

Type 5 · mth Term from the End

5.1

From the End

Find the 10th term from the end of the AP: 3, 8, 13, ..., 253.

Reveal Solution
mth from end = last term − (m−1)d = 253 − (10−1)×5
✓ 10th from end = 208
5.2

From the End

Find the 6th term from the end of the AP: 17, 14, 11, ..., −40.

Reveal Solution
6th from end = −40 − (6−1)×(−3) = −40 + 15
✓ 6th from end = −25
5.3

From the End

Find the 12th term from the end of the AP: 5, 9, 13, ..., 185.

Reveal Solution
12th from end = 185 − (12−1)×4
✓ 12th from end = 141
5.4

From the End

Find the 5th term from the end of the AP: 21, 18, 15, ..., −81.

Reveal Solution
5th from end = −81 − (5−1)×(−3) = −81 + 12
✓ 5th from end = −69
5.5

From the End

Find the 15th term from the end of the AP: 2, 7, 12, ..., 152.

Reveal Solution
15th from end = 152 − (15−1)×5
✓ 15th from end = 82
⚠️ Common mistake: applying the standard a+(n-1)d formula counting from the start instead of using the "from the end" version — the direction of d effectively flips when counting backward from the last term.

Type 6 · Verifying an AP from Its General Term

6.1

Is It an AP?

Is the sequence with nth term aₙ = 3n + 5 an AP? If so, find a and d.

Reveal Solution

Compute the first three terms

a₁=8, a₂=11, a₃=14

Common difference: 11−8=3, and 14−11=3 — constant
✓ Yes, it's an AP. a = 8, d = 3.
6.2

Is It an AP?

Is the sequence with nth term aₙ = 7 − 4n an AP? If so, find a and d.

Reveal Solution

Compute the first three terms

a₁=3, a₂=−1, a₃=−5

Common difference: −1−3=−4, and −5−(−1)=−4 — constant
✓ Yes, it's an AP. a = 3, d = −4.
6.3

Is It NOT an AP?

Is the sequence with nth term aₙ = n² + 1 an AP?

Reveal Solution

Compute the first three terms

a₁=2, a₂=5, a₃=10

Difference: 5−2=3, but 10−5=5 — NOT constant
✓ No, this is not an AP (the difference changes from 3 to 5).
⚠️ A quadratic expression in n (with an n² term) never produces an AP — only a linear expression in n (of the form An+B) does. Spotting the n² term immediately tells you this can't be an AP without even calculating terms.
6.4

Is It an AP?

Is the sequence with nth term aₙ = (2n/3) + 1 an AP? If so, find a and d.

Reveal Solution

Compute the first three terms

a₁=5/3, a₂=7/3, a₃=3

Common difference: 7/3−5/3=2/3, and 3−7/3=2/3 — constant
✓ Yes, it's an AP. a = 5/3, d = 2/3 (an AP doesn't require whole-number terms).
6.5

Is It an AP?

Is the sequence with nth term aₙ = 9 − 5n an AP? If so, find a and d.

Reveal Solution

Compute the first three terms

a₁=4, a₂=−1, a₃=−6

Common difference: −1−4=−5, and −6−(−1)=−5 — constant
✓ Yes, it's an AP. a = 4, d = −5.

Type 7 · Word Problems

7.1

Salary Increment

A person's starting salary is ₹20,000 per year, with an annual increment of ₹2,000. Find their salary in the 10th year.

Reveal Solution

This is a direct nth-term application

a=20000, d=2000

a₁₀ = 20000 + (10−1)×2000
✓ Salary in the 10th year = ₹38,000
7.2

Stadium Seating

In an auditorium, the first row has 20 seats, and each subsequent row has 4 more seats than the one before it. Find the total number of seats in the first 15 rows.

Reveal Solution

This is a sum-of-n-terms application

a=20, d=4, n=15

S₁₅ = 15/2 × [2(20) + (15−1)(4)] = 7.5 × [40+56]
✓ Total seats in first 15 rows = 720
7.3

Loan Repayment

A loan is repaid in monthly instalments. The first instalment is ₹1,000, and each subsequent instalment increases by ₹100. Find the total amount repaid in 12 months.

Reveal Solution

a=1000, d=100, n=12

S₁₂ = 12/2 × [2(1000) + (12−1)(100)] = 6 × [2000+1100]
✓ Total repaid in 12 months = ₹18,600
7.4

Stacked Logs

Logs are stacked so that the bottom row has 20 logs, and each row above has 1 fewer log than the row below, with the top row having exactly 1 log. Find the number of rows and the total number of logs.

Reveal Solution

a=20, d=−1, last term=1. First find n

20+(n−1)(−1)=1 → n=20

S₂₀ = 20/2 × (20+1) = 10 × 21
✓ 20 rows, total logs = 210
7.5

Annual Production

A company produces 800 units in its first year of operation, and increases production by 60 units every following year. Find the production in the 12th year, and the total production over the first 12 years.

Reveal Solution

a=800, d=60

a₁₂ = 800+(12−1)(60) = 800+660. S₁₂ = 12/2 × [2(800)+(12−1)(60)] = 6×[1600+660]
✓ Production in the 12th year = 1,460 units. Total over 12 years = 13,560 units.

Which of These 7 Types Would You Actually Recognise on Sight?

Reading through 35 solved problems and being able to solve fresh ones without the type named for you are different skills. The real exam test is spotting which of these 7 patterns a new question belongs to.

Genelis Performance Map

What a Genelis weak area map looks like after working through Arithmetic Progressions practice

nth term & number of terms
82%
Sum of n terms
70%
Finding AP from two conditions
54%
Word problems — modelling from scratch
33%

Next session: word-problem modelling (33%) — not more direct nth-term drilling. Genelis tracks accuracy by type, not just by topic, so it knows exactly which pattern needs more reps.

Genelis is an AI-powered personalized learning platform built on Adaptive Personalized Intelligence. The Genelis learning system generates fresh, unlabelled Arithmetic Progressions problems across all 7 types, tracks your accuracy on each specifically, and logs every wrong answer to your wrong-question notebook for reattempt.

Step 1 Attempt fresh problems
→
Step 2 Type-level gap detected
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Step 3 AI notes for weak pattern
→
Step 4 Wrong Qs auto-logged
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Step 5 Reattempt that type
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Result Gap closed. Map updates. ✓
Practise unlimited fresh Arithmetic Progressions problems on Genelis — free →
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Frequently Asked Questions

Questions Students Commonly Ask

Quick answers to the most common questions related to this guide.

What question types appear in CBSE Class 10 Arithmetic Progressions?

Seven types recur most often: finding the nth term using the general term formula, finding the number of terms or which term equals a given value, finding the sum of n terms, determining an AP's first term and common difference from two given terms, finding a term counted from the end, verifying whether a given general term expression represents an AP, and word problems modelling real-world scenarios as an AP.

Can finding the number of terms in an AP give two different valid answers?

Yes, in a specific situation: if a term in the sequence happens to equal exactly zero, then the sum up to that term and the sum up to the term just before it are identical, since adding zero doesn't change a total. This means two different values of n can both correctly satisfy a given sum — both are valid answers, and this is a genuine mathematical result, not an error in the working.

How do I find the first term and common difference when given two specific terms of an AP?

Write the general term formula for each given term as a separate equation in terms of the first term (a) and common difference (d), then solve the two equations simultaneously. Subtracting one equation from the other eliminates a and isolates d directly, after which substituting back gives a.

How do I check whether a given expression for the nth term actually represents an arithmetic progression?

Calculate the first three terms by substituting n=1, n=2, and n=3 into the expression, then check whether the difference between consecutive terms is the same both times. If the common difference is constant, the expression represents an AP; if the difference changes between the first pair and the second pair, it does not, regardless of how similar the expression might look to a typical AP formula.

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