Arithmetic Progression and Geometric Progression Formulas

ARITHMETIC PROGRESSION AND GEOMETRIC PROGRESSION FORMULAS

Arithmetic Progression Formulas

General form of an arithmetic progression :

a₁, a₁ + d, a₁ + 2d, a₁ + 3d,.............

a₁ ---> first term

d ----> common difference

Formula to find the common difference :

d = second term - first term

d = a₂ - a₁

Formula to find n^th term of an arithmetic progression :

a_n = a₁ + (n - 1)d

Let us consider the following arithmetic progression.

a₁, a₁ + d, a₁ + 2d, a₁ + 3d,.............l

l ----> last term

Formula to find number of terms 'n' in the above arithmetic progression :

n = [(l - a)/d] + 1

Formula to find sum to 'n' terms of an arithmetic progression :

S_n = (n/2)[2a₁ + (n - 1)d]

S_n = (n/2)[a₁ + l]

If three numbers are in arithmetic progression, then they have to be assumed as

a - d, a, a + d

If four numbers are in arithmetic progression, then they have to be assumed as

a - 3d, a- d, a + d, a + 3d

Special Series Formulas

Sum of first 'n' natural numbers :

1 + 2 + 3 + ......... + n = n(n + 1)/2

Sum of squares of first 'n' natural numbers :

1² + 2² + 3² + ......... + n² = [n(n + 1)(2n + 1)]/6

Sum of cubes of first 'n' natural numbers :

1³ + 2³ + 3³ + ......... + n³ = [n(n + 1)/2]²

Sum of odd numbers :

1 + 3 + 5 + ......... + l = n²

where n = (l + 1)/2

Geometric Progression Formulas

General form of a geometric progression :

a₁, a₁r, a₁r², a₁r³, ............

a₁ ---> first term

r ----> common ratio

Formula to find the common ratio :

r = second term/first term

d = a₂/a₁

Formula to find n^th term of an geometric progression :

a_n = a₁r^{n - 1}

Formula to find sum to 'n' terms of a geometric progression :

When r > 1 :

S_n = a₁(rⁿ - 1)/(r - 1)

When r < 1 :

S_n = a₁(1 - rⁿ)/(1 - r)

When r = 1 :

S_n = na₁

Formula to find sum of infinite geometric progression :

S_∞= a₁/(1 - r)

where -1 < r < 1.

If three numbers are in geometric progression, then they have to be assumed as

a/r, a, ar

If four numbers are in geometric progression, then they have to be assumed as

a/r³, a/r, ar, ar³

Problem 1 :

Find the common difference and 15^th term of the arithmetic progression :

125, 120, 115, 110, ..........

Solution :

Common difference :

d = second term - first term

= 120 - 125

= -5

Formula to find n^th term of an arithmetic progression :

a_n = a₁ + (n - 1)d

Substitute n = 15, a₁ = 125 and d = -5.

a₁₅ = 125 + (15 - 1)(-5)

= 125 + (14)(-5)

= 125 - 70

= 55

Problem 2 :

Find the sum of the terms in the following arithmetic progression :

3 + 7 + 11 + 15 + ........ to 25 terms

Solution :

a₁ = 3

d = second term - first term

= 7 - 3

= 4

Formula to find sum to 'n' terms of an arithmetic progression :

S_n = (n/2)[2a₁ + (n - 1)d]

Substitute n = 25, a₁ = 3 and d = 3.

S₂₅ = (25/2)[2(3) + (25 - 1)(4)]

= (25/2)[2(3) + (24)(4)]

= (25/2)[6 + 96]

= (25/2)[102]

= 25(51)

= 1275

Problem 3 :

Find the number of terms in the following arithmetic progression :

7, 12, 17, ........ 252

Solution :

a₁ = 7

l = 252

d = second term - first term

= 12 - 7

= 5

Formula to find number of terms 'n' in the above arithmetic progression :

n = [(l - a)/d] + 1

Substitute a₁ = 7, l = 252 and d = 5.

n = [(252 - 7)/5] + 1

= [(245)/5] + 1

= 49 + 1

= 50

Problem 4 :

Find the common ratio and 15^th term of the geometric progression :

2, 4, 8, ..........

Solution :

Common ratio :

r = second term/first term

= 4/2

= 2

Formula to find n^th term of a geometric progression :

a_r = a₁r^n-1

Substitute n = 15, a₁ = 2 and r = 2.

a₁₅ = 2(2)^{15 - 1}

= 2(2)¹⁴

= 2(16384)

= 32768

Problem 5 :

Find the sum of the terms in the following geometric progression :

2 + 6 + 18 + 54 + ........ to 10 terms

Solution :

a₁ = 2

r = second term/first term

= 6/2

= 3 > 1

Formula to find sum to 'n' terms of a geometric progression when r > 1 :

S_n = a₁(rⁿ - 1)/( r - 1)

Substitute a₁ = 2, r = 3 and n = 10.

S₁₀ = 2(3¹⁰ - 1)/(3 - 1)

= 2(59049 - 1)/2

= 2(59048)/2

= 59048

Problem 6 :

Divide 69 into three parts which are in arithmetic progression are such that the product of first two parts is 483.

Solution :

Since we divide 69 into three parts which are in arithmetic progression, the three parts of 69 can be assumed as

a - d, a, a + d

Then,

(a - d) + a + (a + d) = 69

a - d + a + a + d = 69

3a = 69

Divide both sides by 3.

a = 23

Given : The product of first two parts is 483.

(a - d)(a) = 483

Substitute a = 23.

(23 - d)(23) = 483

Divide both sides by 23.

23 - d = 21

Subtract 23 from both sides.

- d = -2

Multiply both sides by -1.

d = 2

The three parts of 69 which are in arithmetic progression are

(a - d), a, (a + d)

(23 - 2), 2, (23 + 2)

21, 2, 25

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.