Arithmetic Progression in Maths

To find the 15th term of the arithmetic progression, we use the formula for the n-th term of an AP: a_n ​ = a + (n − 1) ⋅d

Where:

• a is the first term,
• d is the common difference,
• n is the term number.

Substitute these values into the formula:
a₁₅ = 12 + (15 − 1) ⋅ (−3)
a₁₅ = 12 + 14 ⋅ (−3)
a₁₅ = 12 − 42
a₁₅ = −30

The first term (a) is the sum of the first term, so: S₁ = 1² + 2 ⋅ 1 = 1 +2 = 3
Thus, a = 3.
The second term is: a₂ ​ =S₂ − S₁​ = 8 − 3 = 5

The common difference is 5 - 3 = 2

To find the sum of the first n terms (S_n​) of an arithmetic progression, we use the formula: Sn = n/2 ⋅ (2a + (n − 1) ⋅ d)
Substitute the values in the formula:

S₂₀ = 20/2​ ⋅ (2 ⋅ 5 + (20 − 1) ⋅ 3)
S₂₀​ = 10 ⋅ (10 + 19 ⋅ 3)
S₂₀ = 10 ⋅ (10 + 57)
S₂₀ = 10 ⋅ 67
S₂₀ = 670

The n-th term of an arithmetic progression (AP) is given by the formula: a_n = a + (n − 1) ⋅ d

Given:

• The 4th term is 20 (a₄ = 20),
• The 10th term is 44 (a₁₀ = 44).

For the 4th term:

a₄ = a + (4 − 1) ⋅ d
a₄ = a + 3d = 20 ......(1)

For the 10th term:

a₁₀ = a + (10 − 1) ⋅ d
a₁₀ = a + (10 - 1) ⋅ d = 44 ......(2)

Subtract Equation (1) from Equation(2)

(a + 9d) − (a + 3d) = 44 − 20
6d = 24
d = 4

The formula for the n-th term of an arithmetic progression (AP) is: a_n = a + (n − 1) ⋅ d

Here:

• a = x,
• d = 2x.

Substitute these values into the formula:

a_n = x + (n − 1) ⋅ 2x
a_n = x + 2xn − 2x
a_n = 2xn - x

The sum of the first n terms of an AP is given by: S_n = n/2[2a + (n − 1)d

For the first 10 terms S₁₀: S₁₀ = 10/2[2a + (10 − 1)d]
S₁₀ = 5[2a + 9d
150 = 5[2a + 9d]
2a + 9d = 30 .....(1)

For the first 20 terms (S₂₀): S₂₀ ​= 20/2​[2a + (20 − 1) d]
S₂₀ = 10[2a + 19d]
700 = 10[2a + 19d]
2a + 19d = 70 .....(2)

Subtract the two equations:
(2a + 19d) − (2a + 9d) = 70 − 30
10d = 40
d = 4

Determine the smallest multiple of 7 greater than 50: The smallest multiple is 56.

Determine the largest multiple of 7 less than 500: The largest multiple is 497.

Form an AP:
First term (a) = 56,
Common difference (d) = 7,
Last term (l) = 497.

Find the number of terms (n) in the AP using the n-th term formula: a_n = a + (n − 1)d

497 = 56 + (n - 1)7
497 = 56 + 7n - 7
449 = 7n
n = 64

The average of the first (a) and last (l) terms is given by:
(a + l)/2 = 15
a + l = 30 .....(1)

The last term (l) of an AP is given by: l = a + (n − 1)d
Substitute n = 12and d = 3:
l = a + (12 − 1) ⋅ 3 = a + 33 .....(2)

Substitute equation (2) into equation (1):
a + (a + 33) = 30
2a + 33 = 30
2a = −3
a = −1.5

The sum of the first nnn terms (S_n​) of an AP is given by:

S_n = n/2 ⋅ (a + l)

Substitute n = 12 , a = −1.5, and l = a + 33 = −1.5+33 = 31.5:
S₁₂ = 12/2 ⋅ (−1.5 + 31.5)
S₁₂ = 6 ⋅ 30
S₁₂​ = 180

The first sequence a_n: a_n = 3n + 1, n = 1, 2, …, 120

The second sequence b_n: b_n = 4n + 1, n = 1, 2, …, 120

For an element to be common, it must satisfy: a_n = b_m

Substitute the expressions for a_n and b_m​: 3n + 1 =4 m + 1

Simplify:
3n = 4m
n/m = 4/3

Let: n = 4k, m = 3k
where k is a positive integer.

since n and m are indices, both must satisfy their respective ranges:

1. For n = 4k: 1 ≤ 4k ≤ 120 ⇒ 1 ≤ k ≤ 30
2. For m = 3k: 1 ≤3k ≤ 120 ⇒ 1 ≤ k ≤ 40

The smallest common range is:
1 ≤ k ≤ 30
Each k corresponds to one common element, so there are 30 common elements.

• First term (a₁):
  The smallest multiple of 11 greater than or equal to 500 is: a₁ = 506
• Last term (a_n):
  The largest multiple of 11 less than or equal to 5000 is: a_n = 4994
• Number of terms (n):
  The number of terms is: n = (last term − first term)/common difference + 1
  Substituting: n = (4994 − 506)/11 + 1 = 409

S_n​ = 2n ​ ⋅(first term + last term)

Substituting:
S_n = 409/2 ⋅ (506 + 4994)
S_n ​ =409⋅2750 = 1,124,750