SERIES

A series is a partial sum of a sequence. If T₁, T₂ ,T₃ ,T₄ .....are the terms of a sequence, then S = T₁, T₂ ,T₃ ,T₄, ...... is the corresponding series.

If the sequence contains finite number of terms then the corresponding series is finite. If the sequence contains infinite number of terms then the corresponding series is infinite. The sum T₁ + T₂ + T₃ + T₄ ...... +Tₙ of n terms of a sequence T₁ , T₂ ,T₃ ,T₄ ,......Tₙ is designated Sₙ.

Thus,

Sₙ = T₁ + T₂ + T₃ + T₄ + ......Tₙ

Also,

Sₙ₋₁ = T₁ + T₂ + T₃ + T₄ + .....Tₙ₋₁

Note that

Sₙ - Sₙ₋₁ = Tₙ

Example 1: The sum of an AP is 21. If the difference between the 3rd term and the 1st term is 4, find the sum of the next four terms of the sequence.

Solution

S₃ = T₁ + T₂ + T₃ = 21

Let the common difference be d

T₁ = a

T₃ = a + 2d

T₃ - T₁ = 2d

2d = 4

d = 2

T₂ = a + 2

T₃ = a + 4

S₃ = T₁ + T₂ + T₃

21 = a + ( a + 2 ) + ( a + 4 )

21 = 3a + 6

3a = 15

a = 5

Let S be the sum of the next four terms of the sequence, then

S = T₄ + T₅ + T₆ + T₇

T₄ = a + 3d

T₄ = 5 + ( 3 x 2 )

T₄ = 5 + 6

T₄ = 11

T₅ = a + 4d

T₅ = 5 + ( 4 × 2 )

T₅ = 5 + 8

T₅ = 13

T₆ = a + 5d

T₆ = 5 + ( 5 × 2 )

T₆ = 5 + 10

T₆ = 15

T₇ = a + 6d

T₇ = 5 + ( 6 × 2 )

T₇ = 5 + 12

T₇ = 17

S = T₄ + T₅ + T₆ + T₇

S = 11 + 13 + 15 + 17

S = 56

Sum of n terms of an AP

Let the first term of a linear sequence be a and the common difference be d. We wish to find an expression for the sum of n terms of a linear sequence.

Let this sum be Sₙ, then

Sₙ = n / 2 [ 2a + ( n - 1 ) d ]

Example 2: Find the sum of the first twenty terms of the linear sequence 1 , 5 , 9 , 13 ,.....

Solution

a = 1 , d = 4 , n = 20

Sₙ = n / 2 [ 2a + ( n - 1 ) d ]

S₂₀ = 20 / 2 [ 2(1) + ( 20 - 1 ) 4 ]

S₂₀ = 10 [ 2 + 76 ]

S₂₀ = 10 × 78

S₂₀ = 780

Example 3: The sum of the first ten terms of an AP is 225. Find the sum of the next twenty terms of the progression given that the sum of the first twenty terms of the progression is 1010.

Solution

The sum of the first ten terms of the sequence + the sum of the next twenty terms of the sequence = the sum of the first 30 terms of the sequence

Sₙ = n / 2 [ 2a + ( n - 1 ) d ]

S₂₀ = 10 [ 2a + 19d ] = 1010......(1)

S₁₀ = 5 [ 2a + 9d ] = 255 .......(2)

Multiplying (2) by 2

20a + 90d = 510 .......(3)

Subtracting (3) from (1) we have

100d = 1010 - 510

100d = 500

d = 5

From (2)

10a + 45(5) = 255

10a + 225 = 255

10a = 30

a = 3

S₃₀ = 15 [ 2(3) + 29(5) ]

 S₃₀ = 15 [ 6 + 145 ]

S₃₀ = 15 × 151

S₃₀ = 2265

Let S be the sum of the next twenty terms required, then

S = S₃₀ - S₁₀

S = 2265 - 255

S = 2010

Sum of n terms of a GP

Sₙ = a ( 1 - rⁿ ) / ( 1 - r ) .......(1)

Sₙ = a ( rⁿ - 1 ) / ( r - 1 ) .......(2)

The formula (1) is applicable if r < 1 while the formula (2) is  applicable if r > 1

The sum of the n terms as n approaches infinity is called the "sum to infinity" of the series and is designated S∞

Thus

S∞ = a / ( 1 - r )

Example 4 : The third term of a GP is 63 and the fifth term is 567. Find the sum of the first six terms of the progression.

Solution

T₃ = ar² = 63

T₅ = ar⁴ = 567

ar⁴ / ar² = 567 / 63

r² = 9 , r = 3

From ar² = 63

9a = 63 , a = 7

Sₙ = a ( rⁿ ⁻ 1 ) / r - 1

S₆ = 7 ( 3⁶ - 1 ) / 3 - 1

S₆ = 7 / 2 × 728

S₆ = 7 × 364

S₆ = 2548

Example 5: Find the sum of the first 6 terms of the exponential sequence 18 , 6 ,2 , ......

Solution

a = 18 , r = ⅓ , n = 6

Sₙ = a ( 1 - rⁿ ) / 1 - r

S₆ = 18 ( 1 - (⅓)⁶ ) / 1 - ⅓

S₆ = 728 / 27

Example 6: Find the sum to infinity of the sequence

1 , ¼ ,(⅛ × ½) ,(⅛ × ⅛), ...

Solution:

a = 1 , r = ¼

S∞ = a / 1 - r = 1 / 1 - ¼ = 1 / ¾ = 4/3.

See algebraic identities