Arithmetic Progression

Definition:- A sequence of numbers is called an Arithmetic progression if the difference between any two consecutive terms is always the same. For example, 2, 4, 6, 8, 10 is an AP because difference between any two consecutive terms in the series (common difference) is same 4 – 2 = 6 – 4 = 8 – 6 = 10 – 8 = 2. 

Important formulas:- 

Common difference (d) = Difference between any two consecutive terms.

nth term (T_n) = a + (n − 1) × d

Sum of n terms = (n/2)[2a + (n − 1) × d] 

Sum of A.P when the last term of the same A.P is given = (n/2) (first term + last term)

Let’s learn these concepts with the help of some questions.

Question 1: Find the common difference of the series 12, 27, 42, 57, 72……..

Solution:

Given:

The given series is 12, 27, 42, 57, 72……..

Formula used:

Common difference (d) = Difference between any two consecutive terms.

Calculation:

So, d = 27 – 12 = 42 – 27 = 57 – 42 = 72 – 57 = 15

The required common difference is 15.

Question 2: Find the 12th term of the series 8, 12, 16, 20……..

Solution:

Given:

The given series is 8, 12, 16, 20……..

Formula used:

nth term (Tn) = a + (n − 1) × d

Calculation:

Here, a = 8

d = 12 – 8 = 4

And, n = 12

We know that, nth term (T_n) = a + (n − 1) × d

So, T₁₂ = 8 + (12 – 1)3 = 8 + 33 = 41

∴ The required value is 41.

Question 3: Find the number of terms in the series 15, 30, 45, 60, …….., 480.

Solution:

Given:

We have the series 15, 30, 45, 60, …….., 480.

Formula used:

n = {(last term – first term)/d} + 1

Where d = difference between any two consecutive terms of the series.

Given:

d = 30 – 15 = 15

According to the formula, we have

n = {(480 – 15)/15} + 1 = 31 + 1 = 32

∴ The required number of terms is 32.

Question 4: Find the 15th terms of the series 122, 133, 144,……..

Solution:

Given:

The given series is 122, 133, 144,……..

Formula used:

Tn = a + (n – 1)d

Where, n = number of term

a = first term

d = difference between any two consecutive terms.

Calculation:

In the given series, d = 133 – 122 = 11

a = 122

According to the formula, we have

T15 = 122 + (15 – 1) × 11 = 122 + 14 × 11 = 122 + 154 = 276

∴ The required number of term is 276.

Question 5: Find the sum of the series 14, 50, 86, 122, 158, 194, 230, 266, 302.

Solution:

Given:

The given series is 14, 50, 86, 122, 18, 194, 230, 266, 302.

Formula used:

Sn = n/2(first term + last term)

Where, n = number of terms

Calculation:

We have, n = 9

First term = 14, last term = 302

According to the formula, we have

Sum of the series = (9/2)(14 + 302) = (9/2) × 316 = 1422

∴ The sum of the series is 1422.

Question 6: If the sum of n terms of an A.P. is 12000 and the first and last terms are 75 and 125 then find the value of n

Solution:

Given:

The sum of the A. P. is 12000

First-term = 75

Last term = 125

Formula used:

Sum of n terms = (n/2)(first term + last term? 

Calculation:

According to the formula, we have

12000 = (n/2)(75 + 125)

So, 100n = 12000

Hence, n = 120

∴ The required value is 120.

Question 7: Find the sum of the series 20, 38, 56, 74…….. till 20 terms

Given:

The given series is 20, 38, 56, 74…….. till 20 terms.

Formula used:

Sn = (n/2){2a + (n – 1)d}

Where, n = number of terms

a = first term

d = difference between two terms

Calculation:

In the given series, a = 20

d = 38 – 20 = 18

n = 20

Then according to the formula, we have

Sum of the series = (20/2){(2 × 20 + (20 – 1) × 18} = 10(40 + 342) = 3820

∴ The sum of the series is 3820.

Question: 8 Find the sum of the series -5, -10, -15, -20, …………. , -855

Solution:

Given:

The given series is -5, -10, -15, -20, …………. , -855

Here a = -5

Last term = -855

Here d = -10 + 5 = -5

Calculation:

Let the number of terms in the series be n

Then, last term = -5 + (n – 1)(-5)

So, -855 = -5 – (n – 1)5

Then, n = 171

We know that, sum of the series = (n/2)(first term + last term) = (171/2)(-5 – 855) = 171 × -430 = -73530

∴ The required value is -73530.

Question: 9 Find the sum of the first 1000 odd numbers.

Solution:

Given:

The required series is 1, 3, 5, 7………….

Formula used:

Sn = (n/2){2a + (n – 1)d}

Where, n = number of terms

a = first term

d = difference between any two consecutive terms.

Calculation:

Here, a = 1

d = 3 – 1 = 2

And, n = 1000

So, sum of the series = (1000/2){2 + (1000 – 1)2} = 500 × 2000 = 10,00,000.

∴ The required value is 1000000.

Question: 10 Find the sum of the first 30 multiples of 5.

Solution:

Formula used:

Sn = (n/2)[2a + (n − 1) × d] 

Calculation:

The required series is 5, 10, 15, 20……..

Here, a = 5

d = 5

So, Sn = (n/2)[2a + (n − 1) × d] 

= (30/2) (10 + 29 × 5) = 15 × 155 = 2325

∴ The required value is 2325.