Arithmetic Progression Class 10th Important Questions with Answers Mathematics

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

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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

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.

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.