Arithmetic Progressions Olympiad Questions, Download PDF

Arithmetic Progressions Olympiad Questions: Arithmetic progression is an important topic in mathematics. It builds a strong base for higher-level problem solving. Many competitive exams include questions from this topic. Students preparing for Olympiads must practice different types of questions regularly.

Here, we focus on the AP Sequence Questions Olympiad. It will help you understand the pattern of questions asked in exams. You will also find practice sets and solved problems for better clarity.

Arithmetic Progression Olympiad Questions PDF

Students often look for well-structured practice material. A good PDF helps in revision and regular practice. It also allows you to solve questions anytime without distraction.

Below are different sets of Arithmetic Progressions Olympiad Questions. These sets are useful for practice and concept building. These are useful AP questions olympiad level for practice.

Arithmetic Progression Problems with Solutions

Practicing questions is important. But understanding solutions is also important. Below are some selected arithmetic progression problems with solutions.

Problem 1

Find the value of n if the nth term of the sequences
25, 29, 33, 37, … and
3, 4, 6, 9, 13, … are equal.

Solution: First sequence: a = 25, d = 4
Tₙ = 25 + (n − 1) × 4

The second sequence pattern gives n = 12

Problem 2

In an AP, 5 times the 5th term equals 8 times the 8th term. Find the 13th term.

Solution:
5(a + 4d) = 8(a + 7d)
a + 12d = 0
13th term = 0

Problem 3

If the 9th term is zero, find the ratio of the 29th and 19th terms.

Solution:
a + 8d = 0
Ratio = 2 : 1

Problem 4

The sum of three numbers in AP is −3, and the product is 8. Find the sum of squares.

Solution: Numbers: a − d, a, a + d
a = −1
Sum of squares = 21

Problem 5

Find the nth term if the sum of n terms is 2n² + n.

Solution: Tₙ = Sₙ − Sₙ₋₁

Sₙ₋₁ = 2(n − 1)² + (n − 1)
= 2(n² − 2n + 1) + n − 1
= 2n² − 4n + 2 + n − 1
= 2n² − 3n + 1

Tₙ = (2n² + n) − (2n² − 3n + 1)
= 4n − 1

Problem 6

Find the 15th term of an AP where a = 3 and d = 5.

Solution:
Tₙ = a + (n − 1)d
T₁₅ = 3 + 14 × 5
= 3 + 70 = 73

Problem 7

Find the sum of first 20 terms of an AP where a = 2 and d = 3.

Solution: Sₙ = n/2 [2a + (n − 1)d]
S₂₀ = 20/2 [4 + 19 × 3]
= 10 [4 + 57] = 10 × 61 = 610

Problem 8

If the 8th term of an AP is 40 and first term is 5, find d.

Solution: T₈ = a + 7d
40 = 5 + 7d
7d = 35
d = 5

Problem 9

Find the number of terms in AP: 4, 9, 14, …, 99

Solution:
a = 4, d = 5
Tₙ = 99

99 = 4 + (n − 1) × 5
95 = 5(n − 1)
n − 1 = 19
n = 20

Problem 10

Find the sum of AP: 6 + 10 + 14 + … + 102

Solution:
a = 6, d = 4
Last term = 102

102 = 6 + (n − 1) × 4
96 = 4(n − 1)
n = 25

Sₙ = 25/2 [12 + 96]
= 25/2 × 108 = 1350

Problem 11

Find the middle term of AP: 2, 6, 10, …, 98

Solution:
a = 2, d = 4

98 = 2 + (n − 1) × 4
96 = 4(n − 1)
n = 25

Middle term = 13th term
T₁₃ = 2 + 12 × 4 = 50

Problem 12

If Sₙ = n² + 3n, find nth term.

Solution:
Tₙ = Sₙ − Sₙ₋₁

Sₙ₋₁ = (n − 1)² + 3(n − 1)
= n² − 2n + 1 + 3n − 3
= n² + n − 2

Tₙ = (n² + 3n) − (n² + n − 2)
= 2n + 2

Problem 13

Find the common difference if the 6th term is 25 and the 2nd term is 9.

Solution:
T₆ = a + 5d = 25
T₂ = a + d = 9

Subtract:
4d = 16
d = 4

Problem 14

Find three numbers in AP whose sum is 21 and product is 315.

Solution: Let numbers be a − d, a, a + d

3a = 21 → a = 7

Numbers: 7 − d, 7, 7 + d

Product = 315

(7 − d)(7)(7 + d) = 315
7(49 − d²) = 315
49 − d² = 45
d² = 4 → d = 2

Numbers = 5, 7, 9

Problem 15

Find the sum of the first 30 natural numbers using the AP formula.

Solution:
a = 1, d = 1

Sₙ = n/2 [2a + (n − 1)d]
S₃₀ = 30/2 [2 + 29]
= 15 × 31 = 465

Arithmetic progression is an important and scoring topic. With regular practice, students can improve accuracy and speed.