ARITHMETIC SERIES IN SIGMA NOTATION

Example 1 : 

Evaluate :

_{r = 1}∑^{r = 10} (2r + 7)

Solution :

Substitute the first few values into the formula. 

_{r = 1}∑^{r = 10} (2r + 7) = 2(1) + 7 + 2(2) + 7 + 2(3) + 7 + ......

= 9 + 11 + 13 + ......

The above is an arithmetic series with a₁ = 9 and d = 2.

Use S_n = (n/2)[2a₁ + (n - 1)d] to find the sum. 

= (10/2)[2(9) + (10 - 1)2]

= 5[18 + (9)2]

= 5[18 + 18]

= 5[36]

= 180

Example 2 : 

Evaluate :

_{r = 3}∑^{r = 10} (5r + 2)

Solution :

Substitute the first few values into the formula. 

_{r = 3}∑^{r = 10} (5r + 2) = 5(3) + 2 + 5(4) + 2 + 5(5) + 2 + ......

= 17 + 22 + 27 + ......

The above is an arithmetic series with a₁ = 17 and d = 5.

Use S_n = (n/2)[2a₁ + (n - 1)d] to find the sum (from r = 3 to r = 10, there are 8 terms, so n = 8).

= (8/2)[2(17) + (8 - 1)5]

= 4[34 + (7)5]

= 4[34 + 35]

= 4[69]

= 276

Example 3 : 

Write the following arithmetic series in sigma notation. 

4 + 7 + 10 + 13 + ........... + 28

Solution :

In the given arithmetic series,

a₁ = 4

d = 7 - 4 = 3

Find the formula for n^th term. 

a_n = a₁ + (n - 1)d

Substitute a₁ = 4 and d = 3.

a_n = 4 + (n - 1)3

= 4 + 3n - 3

= 3n + 1

Let a_n = 28.

a_n = 28

3n + 1 = 28

3n = 27

n = 9

In the given arithmetic series, 4 is the 1^st term and 28 is the 9^th term. 

Then, 

4 + 7 + 10 + 13 + ........... + 28 = _{n = 1}∑^{n = 9} (3n + 1)

Example 4 : 

Write the following arithmetic series in sigma notation. 

-1 + 3 + 7 + 11 + ........... + 55

Solution :

In the given arithmetic series,

a₁ = -1

d = 3 - (-1) = 3 + 1 = 4

Find the formula for n^th term. 

a_n = a₁ + (n - 1)d

Substitute a₁ = -1 and d = 4.

a_n = -1 + (n - 1)4

= -1 + 4n - 4

= 4n - 5

Let a_n = 55.

a_n = 55

4n - 5 = 55

4n = 60

n = 15

In the given arithmetic series, -1 is the 1^st term and 55 is the 15^th term. 

Then, 

-1 + 3 + 7 + 11 + ........... + 55 = _{n = 1}∑^{n = 15} (4n - 5)

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