CBSE X Maths — Arithmetic Progressions: Notes, Practice Questions & Worked Solutions

CBSE Class X Mathematics (041/241) · Unit II: Algebra

Arithmetic Progressions

An arithmetic progression (AP) is a list of numbers where each term increases by the same fixed amount. Once you know the first term and that fixed difference, you can find any term — or the sum of any number of terms — directly.

1. The nth Term and Sum of an AP

aₙ = a + (n − 1)d

Sₙ = n/2 [2a + (n − 1)d]

where a is the first term, d is the common difference, and n is the number of terms.

Worked example — finding a term Find the 10th term of the AP 2, 7, 12, …
a = 2, d = 5. a₁₀ = 2 + (10−1)(5) = 2 + 45 = 47

Worked example — finding the sum Find the sum of the first 20 terms of the AP 3, 7, 11, …
a = 3, d = 4, n = 20. S₂₀ = 20/2 [2(3) + 19(4)] = 10[6 + 76] = 10(82) = 820

Worked example — finding a and d from two terms If a₃ = 15 and a₉ = 39, find a and d, and hence a₂₀.
a + 2d = 15 …(1) and a + 8d = 39 …(2). Subtracting (1) from (2): 6d = 24 → d = 4. Then a = 15 − 8 = 7.
a₂₀ = 7 + 19(4) = 7 + 76 = 83

Worked example — how many terms give a certain sum? How many terms of the AP 9, 17, 25, … give a sum of 636?
a=9, d=8. Sₙ = n/2[2(9)+(n−1)8] = 636 → n(10+8n) = 1272 → 4n²+5n−636=0.
Discriminant = 25 + 10176 = 10201 = 101². n = (−5+101)/8 = 12

Common mistakes to watch for

Using n instead of (n−1) in the nth term formula — the first term corresponds to n=1, where (n−1)=0.
Forgetting that d can be negative (a decreasing AP) — always compute d = a₂ − a₁ carefully, including its sign.
When finding the kth term from the end, forgetting to convert it to the correct term-from-the-start position first.

Key Vocabulary Recap

Common difference (d) — the fixed amount added to get the next term

nth term (aₙ) — the general formula for any term in the sequence

Syllabus Reference

This note covers CBSE Class X Mathematics Unit II, Arithmetic Progressions: motivation for AP, derivation of the nth term and sum of the first n terms, and applications in solving daily-life problems.

Practice questions

CBSE Class X Mathematics · Unit II: Arithmetic Progressions — Practice Question Bank (no answers)

Arithmetic Progressions — Practice

Covers: the nth term, sum of n terms, and real-life applications. 10 questions.

Find the 10th term of the AP 2, 7, 12, …

Find the sum of the first 20 terms of the AP 3, 7, 11, …

Which term of the AP 21, 18, 15, … is −81?

Find the sum of the AP 5, 9, 13, … up to 15 terms.

If the 3rd term of an AP is 15 and the 9th term is 39, find the first term and common difference, and hence find the 20th term.

How many terms of the AP 9, 17, 25, … must be taken to give a sum of 636?

Find the sum of all multiples of 7 lying between 100 and 300.

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

Find three numbers in AP whose sum is 24 and product is 440.

Q10

Find the 11th term from the end of the AP 10, 7, 4, …, −62.

More Practice

Q11

Find the sum of the first 51 terms of an AP whose 2nd and 3rd terms are 14 and 18.

Q12

If the 9th term of an AP is zero, prove that its 29th term is twice its 19th term.

Q13

Find the sum: (−5) + (−8) + (−11) + … + (−230).

Q14

If Sₙ = 3n² + 5n for an AP, find the first term and common difference.

Q15

How many terms of the AP 63, 60, 57, … must be taken so that their sum is 693?

Q16

Find the sum of the first 40 positive integers divisible by 6.

Q17

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 AP.

Q18

Find the middle term of the AP 6, 13, 20, …, 216.

Q19

Determine the AP whose 3rd term is 5 and 7th term is 9.

Q20

Find the number of terms of the AP 18, 15½, 13, …, −47, and find the sum of all its terms.

Worked solutions

Show worked solutions

CBSE Class X Mathematics · Unit II: Arithmetic Progressions — Solutions

Arithmetic Progressions — Practice: Solutions

10th term of 2,7,12,…

aₙ=a+(n−1)d, a=2,d=5

2+9(5)

✓47

Sum of first 20 terms of 3,7,11,…

Sₙ=n/2[2a+(n−1)d], a=3,d=4,n=20

10[6+76]

✓820

Which term of 21,18,15,… is −81?

a=21, d=−3. Set aₙ=−81

21+(n−1)(−3)=−81 → (n−1)=34

✓n = 35

Sum of 5,9,13,… up to 15 terms.

a=5,d=4,n=15

S=15/2[10+56]=15/2(66)

✓495

a₃=15, a₉=39. Find a, d, a₂₀.

a+2d=15, a+8d=39. Subtract

6d=24 → d=4, a=15−8=7

a₂₀=a+19d

7+76

✓a = 7, d = 4, a₂₀ = 83

Terms of 9,17,25,… for sum 636.

a=9,d=8. Set up the equation

n/2[18+8(n−1)]=636 → 4n²+5n−636=0

Discriminant=25+10176=10201=101². Solve

n=(−5+101)/8

✓n = 12

Sum of multiples of 7 between 100 and 300.

First multiple ≥100 is 105, last ≤300 is 294. Find n

n=(294−105)/7+1=28

Sum=n/2(first+last)

14×(105+294)=14×399

✓5586

S₇=49, S₁₇=289. Find Sₙ.

From S₇=49: a+3d=7. From S₁₇=289: a+8d=17. Subtract

5d=10 → d=2, a=1

Sₙ=n/2[2a+(n−1)d]=n/2[2+2(n−1)]=n/2(2n)

=n²

✓Sₙ = n²

Three numbers in AP, sum 24, product 440.

Let numbers be a−d, a, a+d. Sum=3a=24

a=8

Product: (8−d)(8)(8+d)=440 → 64−d²=55

d²=9 → d=3

✓The numbers are 5, 8, and 11

Q10

11th from end of 10,7,4,…,−62.

a=10,d=−3. Find total number of terms n from aₙ=−62

10+(n−1)(−3)=−62 → n=25

11th from the end = (25−11+1)=15th term from the start

a₁₅=10+14(−3)=10−42

✓−32

More Practice — Solutions

Q11

S₅₁, a₂=14, a₃=18.

d=18−14=4, a=14−4=10

S₅₁=51/2[2(10)+50(4)]

✓5610

Q12

a₉=0. Prove a₂₉=2×a₁₉.

a+8d=0 → a=−8d. Find a₁₉ and a₂₉ in terms of d

a₁₉=a+18d=10d; a₂₉=a+28d=20d

✓a₂₉=20d=2×10d=2×a₁₉ — proved

Q13

Sum (−5)+(−8)+…+(−230).

a=−5,d=−3. Find n from aₙ=−230

−5+(n−1)(−3)=−230 → n=76

S₇₆=76/2[2(−5)+75(−3)]

38[−10−225]

✓−8930

Q14

Sₙ=3n²+5n. Find a, d.

a₁=S₁=3+5=8. a₂=S₂−S₁=(12+10)−8

a₂=14

✓First term a=8, common difference d=14−8=6

Q15

Terms of 63,60,57,… for sum 693.

a=63,d=−3. Set up Sₙ=693

n²−43n+462=0 → discriminant=1849−1848=1

✓n=21 or n=22 (both valid, since later terms turn negative and reduce the sum back)

Q16

Sum of first 40 multiples of 6.

a=6,d=6,n=40. Last term=6×40=240

S=40/2(6+240)

✓4920

Q17

a₄+a₈=24, a₆+a₁₀=44. Find the AP.

a+3d+a+7d=24 → a+5d=12. a+5d+a+9d=44 → a+7d=22. Subtract

2d=10 → d=5, a=12−25=−13

✓The AP is −13, −8, −3, 2, …

Q18

Middle term of 6,13,20,…,216.

a=6,d=7. Find n: 6+(n−1)7=216

n=31 (odd, so a single middle term = the 16th term)

✓a₁₆=6+15(7)=111

Q19

a₃=5, a₇=9. Find the AP.

a+2d=5, a+6d=9. Subtract

4d=4 → d=1, a=3

✓The AP is 3, 4, 5, 6, …

Q20

Terms of 18,15½,13,…,−47; find n and sum.

a=18,d=−2.5. Find n from aₙ=−47

18+(n−1)(−2.5)=−47 → n=27

S₂₇=27/2[2(18)+26(−2.5)]

27/2[36−65]

✓n = 27 terms, Sum = −391.5

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