Arithmetic Progression Class 10th Important Questions with Answers Mathematics
Updated on June 1, 2025 | By Learnzy Academy
Q1. The sum of first 16 terms of the AP: 10, 6, 2,… is
Q2. If a train travels 5 km in the first minute and increases speed by 2 km per minute, how much distance will it cover in 20 minutes?
Q3. A ladder has rungs placed at equal distances. If the first rung is at 15 cm and the last at 120 cm, with 10 rungs, find the distance between two consecutive rungs.
Q4. A person saves ₹500 in the first month and increases savings by ₹100 every month. How much does he save in the 12th month?
Q5. Which term of the AP: 21, 42, 63, 84,… is 210 ?
Q6. If the 2nd term of an AP is 13 and the 5th term is 25, then its 7th term is
Q7. The sum of the first five multiples of 3 is:
Q8. 20th term from the last term of the A.P. 3, 8, 13, …, 253 is:
Q9. The number of multiples of 4 between 10 and 250 is:
Q10. If 17th term of an A.P. exceeds its 10th term by 7. The common difference is:
Q11. The 21st term of AP whose first two terms are -3 and 4 is:
Q12. Which term of the A.P. 3, 8, 13, 18, … is 78?
Q13. If the sum of three numbers in an A.P. is 9 and their product is 24, then numbers are
Q14. If p, q, r and s are in A.P. then r – q is
Q15. The sum of all two digit odd numbers is
Q16. Find the sum of all natural numbers that are less than 100 and divisible by 4.
Step 1: List the numbers less than 100 that are divisible by 4:
4, 8, 12, 16, ..., 96
This is an arithmetic sequence where:
First term (a) = 4
Last term (l) = 96
Common difference (d) = 4
Step 2: Find the number of terms (n)
Use the formula for the nth term of an arithmetic sequence:
l = a + (n - 1) * d
96 = 4 + (n - 1) * 4
96 = 4n
n = 96 / 4 = 24
Step 3: Find the sum of the sequence
Use the sum formula:
Sum = (n / 2) * (a + l)
Sum = (24 / 2) * (4 + 96)
Sum = 12 * 100
Sum = 1200
Hence the sum of all natural numbers that are less than 100 and divisible by 4 is 1200
Q17. If seven times the 7th term of an A.P. is equal to eleven times the 11th term, then what will be its 18th term?
Step 1: Use the formula for the nₜₕ term of an A.P.
General formula:
Tₙ = a + (n - 1) * d
So:
T₇ = a + 6d
T₁₁ = a + 10d
Step 2: Use the given condition
7 × T₇ = 11 × T₁₁
Substitute the values:
7 × (a + 6d) = 11 × (a + 10d)
=> 7a + 42d = 11a + 110d
=> 7a - 11a + 42d - 110d = 0
=> -4a - 68d = 0
Divide by -4:
=> a + 17d = 0
=> a = -17d ----------------------(1)
Step 5: Find the 18ₜₕ term
T₁₈ = a + 17d
Substitute a = -17d from equation (1)
T₁₈ = -17d + 17d = 0
Hence the 18ₜₕ term is 0.
Q18. The 4ᵗʰ term of an A.P. is zero. Prove that the 25ᵗʰ term of the A.P. is three times its 11ᵗʰ term.
Step 1: Use the formula for the nᵗʰ term of an A.P.
The general formula is:
Tₙ = a + (n - 1) * d
Where:
a = first term
d = common difference
Tₙ = nᵗʰ term
Step 2: Use the given condition
The 4ᵗʰ term is zero:
T₄ = 0
=> a + (4 - 1) * d = 0
=> a + 3d = 0
=> a = -3d
Step 3: Find T₂₅ and T₁₁
T₂₅ = a + (25 - 1) * d = a + 24d
Substitute a = -3d:
T₂₅ = -3d + 24d = 21d
T₁₁ = a + (11 - 1) * d = a + 10d
Substitute a = -3d:
T₁₁ = -3d + 10d = 7d
Step 4: Prove the required condition
T₂₅ = 3 × T₁₁
21d = 3 × 7d = 21d
Since both sides are equal, the statement is true.
Which means the 25ᵗʰ term is three times the 11ᵗʰ term. Hence, proved.
Q19. In an A.P., if the 6th and 13th terms are 35 and 70 respectively, find the sum of its first 20 terms.
Step 1: Use the formula for the nᵗʰ term of an A.P.
The formula for the nᵗʰ term of an A.P. is:
Tₙ = a + (n - 1) * d
Where:
a = first term
d = common difference
Tₙ = nᵗʰ term
Step 2: Set up equations for T₆ and T₁₃
For the 6ᵗʰ term (T₆):
T₆ = 35
=> a + (6 - 1) * d = 35
=> a + 5d = 35 ----------------------- (1)
For the 13ᵗʰ term (T₁₃):
T₁₃ = 70
=> a + (13 - 1) * d = 70
=> a + 12d = 70 --------------------- (2)
Step 3: Solve the system of equations
Now, subtract Equation 1 from Equation 2 to eliminate a:
(a + 12d) - (a + 5d) = 70 - 35
7d = 35
d = 5
Step 4: Find the value of a
Substitute d = 5 into Equation 1:
a + 5(5) = 35
a + 25 = 35
a = 35 - 25 = 10
Step 5: Find the sum of the first 20 terms
The sum of the first n terms of an A.P. is given by the formula:
Sₙ = (n / 2) * [2a + (n - 1) * d]
For the first 20 terms (n = 20):
S₂₀ = (20 / 2) * [2(10) + (20 - 1) * 5]
S₂₀ = 10 * [20 + 95]
S₂₀ = 10 * 115 = 1150
Hence the sum of the first 20 terms of the A.P. is 1150.
Q20. How many multiples of 4 lie between 10 and 250? Also find their sum.
Step 1: Find the first multiple of 4 greater than 10.
Divide 10 by 4:
10 ÷ 4 = 2.5
So, the next integer greater than 2.5 is 3. Therefore, the first multiple of 4 greater than 10 is:
4 × 3 = 12
Thus, the first multiple of 4 greater than 10 is 12.
Step 2: Find the last multiple of 4 less than 250.
Divide 250 by 4:
250 ÷ 4 = 62.5
The next integer smaller than 62.5 is 62. Therefore, the last multiple of 4 less than or equal to 250 is:
4 × 62 = 248
Thus, the last multiple of 4 less than or equal to 250 is 248.
Step 3: Count the multiples of 4 between 10 and 250.
The multiples of 4 form an arithmetic progression:
12, 16, 20, 24, ..., 248
Where:
First term a = 12
Common difference d = 4
Last term l = 248
We use the formula for the nth term of an arithmetic progression:
Tₙ = a + (n - 1) × d
Set Tₙ = 248
Tₙ =248 (the last term)
=> 12 + (n - 1) × 4 = 248
248 - 12 = (n - 1) × 4
236 = (n - 1) × 4
n - 1 = 236 ÷ 4 = 59
n = 60
So, there are 60 multiples of 4 between 10 and 250.
Step 4: Find the sum of these multiples.
The sum of the first n terms of an arithmetic progression is given by the formula:
Sₙ = (n / 2) × (a + l)
For the first 60 terms:
S₆₀ = (60 / 2) × (12 + 248)
S₆₀ = 30 × 260 = 7800
Hence there are 60 multiples of 4 between 10 and 250. And the sum of these multiples is 7800.
Q21. Find the value of m so that m + 2, 4m – 6 and 3m – 2 are three consecutive terms of an AP.
To solve this problem using the concept of common difference in an arithmetic progression (AP), let's proceed step-by-step.
We know the three terms of the AP are:
a₁ = m + 2
a₂ = 4m - 6
a₃ = 3m - 2
In an arithmetic progression, the difference between consecutive terms is constant, i.e.,
a₂ - a₁ = a₃ - a₂
Step 1: Calculate the common difference using the first two terms
a₂ - a₁ = (4m - 6) - (m + 2)
=> a₂ - a₁ = 4m - 6 - m - 2 = 3m - 8
Step 2: Calculate the common difference using the last two terms
a₃ - a₂ = (3m - 2) - (4m - 6)
=> a₃ - a₂ = 3m - 2 - 4m + 6 = -m + 4
Step 3: Set the two common differences equal to each other
3m - 8 = -m + 4
=> 3m + m = 4 + 8
=> 4m = 12
=> m = 3
Hence the value of m is 3.
Q22. Which term of the AP : 21, 18, 15, . . . is – 81? Also, is any term 0? Give reason for your answer.
Given AP: 21, 18, 15, ...
First term a = 21
Common difference d = 18 − 21 = −3
Which term is −81?
Use the formula for the nth term of an AP:
aₙ = a + (n − 1) × d
Substitute values:
−81 = 21 + (n − 1)(−3)
−81 = 21 − 3(n − 1)
−81 = 21 − 3n + 3
−81 = 24 − 3n
−105 = −3n
n = 35
So, −81 is the 35th term of the AP (a₃₅).
Is any term 0?
Set aₙ = 0
0 = 21 + (n − 1)(−3)
0 = 21 − 3(n − 1)
−21 = −3(n − 1)
n − 1 = 7
n = 8
Yes, the 8th term (a₈) is 0.
Q23. Check whether – 150 is a term of the AP : 11, 8, 5, 2 . . .
Given AP: 11, 8, 5, 2, ...
First term a = 11
Common difference d is = 8 − 11 = –3
We are asked: Is –150 a term of this AP?
Use the nth term formula:
aₙ = a + (n − 1) × d
Substitute values:
−150 = 11 + (n − 1)(−3)
−150 = 11 − 3(n − 1)
−150 = 11 − 3n + 3
−150 = 14 − 3n
−164 = −3n
n = 54.67
Since n = 54.67 is not a whole number,
So –150 is not a term of the AP.
Q24. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73
We are given:
11th term (a₁₁) = 38
16th term (a₁₆) = 73
We need to find the 31st term (a₃₁).
As the nth term formula
aₙ = a + (n − 1) × d
From the 11th term:
a + 10d = 38 ------------------- (1)
From the 16th term:
a + 15d = 73 ---------------------(2)
Subtract Equation 1 from Equation 2
(a + 15d) − (a + 10d) = 73 − 38
5d = 35
d = 7
Put d = 7 into Equation (1)
a + 10 × 7 = 38
a + 70 = 38
a = 38 − 70 = −32
Find the 31st term
a₃₁ = a + 30d
a₃₁ = −32 + 30 × 7 = −32 + 210 = 178
Hence the 31st term is 178.
Q25. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
We are given:
Total number of terms = 50
3rd term a₃ = 12
Last term a₅₀ = 106
As aₙ = a + (n − 1) × d
From a₃ = 12:
a + 2d = 12 ------------------------ (1)
From a₅₀ = 106:
a + 49d = 106 --------------------------(2)
Subtract Equation 1 from Equation 2
(a + 49d) − (a + 2d) = 106 − 12
47d = 94
d = 2
Put d = 2 into Equation 1
a + 2 × 2 = 12
a + 4 = 12
a = 8
Now find the 29th term
a₂₉ = a + 28d
a₂₉ = 8 + 28 × 2 = 8 + 56 = 64
Hence the 29th term is 64.
Q26. If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero?
We are given:
3rd term a₃ = 4
9th term a₉ = –8
Need to find which term of the AP is 0
As aₙ = a + (n − 1) × d
From a₃ = 4:
a + 2d = 4 ------------------------- (1)
From a₉ = –8:
a + 8d = –8 -------------------------- (2)
Subtract Equation 1 from Equation 2
(a + 8d) − (a + 2d) = –8 − 4
6d = –12
d = –2
Put d = –2 into Equation 1
a + 2(–2) = 4
a – 4 = 4
a = 8
Find the term where aₙ = 0
0 = a + (n − 1) × d
0 = 8 + (n − 1)(–2)
0 = 8 − 2(n − 1)
−8 = −2(n − 1)
4 = n − 1
n = 5
Hence the 5th term of the AP is 0.
Q27. The 17th term of an AP exceeds its 10th term by 7. Find the common difference
We are given:
The 17th term exceeds the 10th term by 7
That means: a₁₇ − a₁₀ = 7
As aₙ = a + (n − 1) × d
So,
a₁₇ = a + 16d
a₁₀ = a + 9d
a₁₇ − a₁₀ = 7 (Given)
=> (a + 16d) − (a + 9d) = 7
=> a − a + 16d − 9d = 7
=> 7d = 7
=> d = 1
Hence the common difference d is 1
Q28. Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term?
Given AP: 3, 15, 27, 39, ...
First term a = 3
Common difference d = 15 − 3 = 12
We are told that a certain term is 132 more than the 54th term.
So, aₙ = a₅₄ + 132
=> a + (n − 1) × d = a + 53 × d + 132
=> (n − 1) × d = 53 × d + 132
=> (n − 1) × 12 = 53 × 12 + 132
=> (n − 1) × 12 = 636 + 132 = 768
=> (n − 1) = 768 ÷ 12 = 64
=> n = 64 + 1 = 65
Hence the 65th term is 132 more than the 54th term.
Q29. Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
We are given:
Two APs have the same common difference
The difference between their 100th terms is 100
We are asked to find the difference between their 1000th terms
Let the first terms of the two APs be:
First AP: a
Second AP: b
Common difference: d (same for both)
The 100th term of the first AP = a + 99d
The 100th term of the second AP = b + 99d
According to question:
(a + 99d) − (b + 99d) = 100
=> a − b = 100
=> (a + 999d) − (b + 999d) = 100
Hence the difference between their 1000th terms is 100.