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.