Sum of First n Terms of an Arithmetic Sequence

Example 1 :

Find the sum of first 28 terms of the following arithmetic sequence.

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

Solution :

From the given information,

t₁ = 7

d = t₂ - t₁

= 12 - 7

= 5

n = 28

Formula for sum of first n terms of an arithmetic sequence :

S_n = (n/2)[2t₁ + (n - 1)d]

Substitute n = 28, t₁ = 7 and d = 5.

S₂₈ = (28/2)[2(7) + (28 - 1)(5)]

= 14[14 + 27(5)]

= 14[14 + 135]

= 14(149)

= 2086

Example 2 :

Evaluate the following arithmetic series.

5 + 9 + 13 + ......... + 201

Solution :

From the given information,

t₁ = 5

d = t₂ - t₁

= 9 - 5

= 4

n = [(l - t₁)d] + 1

= [(201 - 5)/4] + 1

= 196/4 + 1

= 49 + 1

= 50

Formula for sum of first n terms of an arithmetic sequence (when the last term is known) :

S_n = (n/2)[t₁ + l]

Substitute n = 50, t₁ = 5 and d = 4.

S₅₀ = (50/2)[5 + 201]

= 25[206]

= 5150

Example 3 :

Find the sum of first 28 terms of an arithmetic sequence whose n^th term is 4n -3.

Solution :

t_n  =  4n - 3

Common difference :

d = a₂ - a₁

= 5 - 1

Formula for sum of first n terms of an arithmetic sequence :

S_n = (n/2)[2t₁ + (n - 1)d]

Substitute n = 28, t₁ = 1 and d = 4.

  = (28/2)[2(1) + (28 - 1)4]

= 14[2 + 27(4)]

= 14 [2 + 108]

= 14(110)

 = 1540

Example 4 :

The sum of first n terms of a certain series is given as 2n² -3n . Show that the series is an arithmetic sequence.

Solution :

Given that :

S_n = 2n² - 3n

To get the sum of first 1 term, substitute n = 1 in S_n.

S₁ = 2(1)² - 3(1)

= 2(1) - 3

= 2 - 3

= -1

The sum of first 1 term is nothing but the 1^st term of the series. 

So, the 1^st term -1. That is

t₁ = -1

To get the sum of first 2 terms, substitute n = 2 in S_n.

S₂ = 2(2)² - 3(2)

= 2(4) - 6

= 8 - 6

= 2

Sum of first 2 terms = 2

t₁ + t₂ = 2

Substitute t₁ = -1. 

-1 + t₂ = 2

t₂ = 3

To get the sum of first 3 terms, substitute n = 3 in S_n.

S₃ = 2(3)² - 3(3)

= 2(9) - 9

= 18 - 9

= 9

Sum of the first 3 terms = 9

t₁ + t₂ + t₃ = 2

Substitute t₁ = -1 and t₂ = 3. 

-1 + 3 + t₃ = 9

2 + t₃ = 9

t₃ = 7

The series is 

Sn = t₁ + t₂ + t₃ + .........

= -1 + 3 + 7 + .........

Clearly this is an arithmetic series with t₁ = -1 and d = 4.

Example 5 :

The 104^th term and 4^th term of an arithmetic sequence are 125 and 0 respectively. Find the sum of first 35 terms.

Solution :

Solve (1) and (2).

(1) - (2) :

(t₁ + 103d) - (t₁ + 3d) = 125 - 0

t₁ + 103d - t₁ - 3d = 125

100d = 125

d = 5/4

Substitute d = 5/4 in (2). 

t₁ + 3(5/4) = 0

t₁  + 15/4 = 0

t₁ = -15/4

Formula for sum of first n terms of an arithmetic sequence :

S_n  =  (n/2)[2t₁ + (n - 1)d]

Substitute n = 35, t₁ = -15/4 and d = 5/4.

S₃₅ = (35/2)[2(-15/4) + (35 - 1)(5/4)]

= (35/2)[2(-15/4) + 34(5/4)]

= (35/2)[-15/2 + 17(5/2)]

= (35/2)[-15/2 + 85/2]

= (35/2)[70/2]

= (35/2)[35]

= 1225/2

= 612.5

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