Sum To n Terms of an Arithmetic Progression

The sum of the first n terms of an AP is given by

Note 1: In the above formula, there are four quantities viz. S_n, a, n and d. If any three of these are known, the fourth can be determined. Sometimes two of these quantities are given, in such cases remaining two quantities are provided by some other relation.

Note 2: If the sum S_n of n terms of a sequence is given, then nth term a_n of the sequence can be determined by the following formula.

a_n = S_n – S_n-1

Example 1: Find the sum of 10 terms of the A.P. 1, 4, 7, 10 ….

Solution: Here values of a (first term) = 1, d (common difference) = 3, n (number of terms to be added) = 10

We know that the sum of first n terms of the Arithmetic Progress, whose first term = a and common difference = d is

Example 2: If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20th term.

Solution: Here S₁₄ (sum of first 14 term) = 1050

n (number of terms added) = 14

a (first term) =10

We know that the sum of first n terms of the Arithmetic Progress, whose first term = a and common difference = d is

Example 3: How many terms of the AP: 24, 21, 18 . . . must be taken so that their sum is 78?

Solution: Here, a = 24, d = 21 – 24 = –3, S_n = 78. We need to find n.

We know that the sum of first n terms of the Arithmetic Progress, whose first term = a and common difference = d is

This implies the sum of the first 4 terms = the sum of the first 13 terms = 78.

Both values of ‘n’ are acceptable. So, the number of terms is either 4 or 13. This is because the sum of the terms from 5th to 13th will be zero as ‘a’ is positive and ‘d’ is negative that makes some terms positive and some negative which will cancel out each other.

Example 4: Find the sum of the first 100 positive integers.

Solution: Let S = 1 + 2 + 3 + . . . + 100

Using the formula  for the sum of the first n terms of an AP, we have

So, the sum of the first 100 positive integers is 5050.

Example 5: Find the sum of first 20 terms of an A.P. whose nth term is given by a_n = 2 + 3n

Solution: We have, a_n = 2 + 3n

Therefore,

a₁ = 2 + 3x1 = 5

a₂ = 2 + 3x2 = 8

a₃ = 2 + 3x3 =11 and so on

d = a₂-a₁ = 8-5 =3

To find S₂₀, we have a = 5, d = 3, n = 20.

We know that the sum of first n terms of the Arithmetic Progress, whose first term = a and common difference = d is

So, the sum of first 20 terms is 670.

Example 6: A manufacturer of electric car produced 500 cars in the second year and 700 cars in the sixth year. Assuming that the production increases uniformly by a fixed number every year, find the production in the 1st year and the total production in first 6 years

Solution: Here, a₂ = 500 and a₆ = 700

Or we write a + d = 500 and a + 5d = 700

Therefore, production of electric car in the first year is 50.

We know that the sum of first n terms of the Arithmetic Progress, whose first term = a and common difference = d is

Thus, the total production of electric car in first 6 years is 1050

Example 7: Find the sum of all natural numbers between 500 and 1000 which are exactly divisible by 3.

Solution: Clearly, the numbers between 250 and 1000 which are divisible by 3 are 501, 504, 507 ….. 999. This is an A.P. with first term a = 501, common difference = 3 and last term = 999. Let there be n terms in this A.P. Then,