EXAMPLES OF FINDING THE SUM OF AN ARITHMETIC SERIES

Question 1 :

Find the sum of the series

34 + 32 + 30 + .................. + 10

Solution :

S_n = (n/2) [a + l]

a = 34, d = 32 - 34  =  -2, l = 10 = a n

a_n  =  a + (n - 1) d

10  =  34 + (n - 1) (-2)

10 - 34  =  (n - 1) (-2)

-24  =  (n - 1) (-2)

 -24/(-2)  =  (n - 1)

12  =  n - 1

n  =  12 + 1

n  =  13

Hence the number of terms in the sequence is 13.

S₁₃  =  (13/2) [34 + 10]

 S₁₃  =  (13/2) [ 44]

  =  13(22)

  =  286

Question 2 :

Find the sum of the series

- 5 + (-8) + (-11) + .............+ (-230)

Solution :

S n = (n/2) [a + l]

 a = -5    d = -8 - (-5) ==> -3, l = -230 = a_n

a_n  =  a + (n - 1) d

-230  =  -5 + (n - 1) (-3)

-230 + 5  =  (n - 1) (-3)

-225  =  (n - 1) (-3)

 225/3  =  (n - 1)

75  =  n - 1

   n  =  75 + 1

   n  =  76

Hence the number of terms in the sequence is 76.

S₇₆ = (76/2) [-8 + (-5)]

 S₇₆ = 38 [ -8 - 5]

  =  38 (-13)

  =  -494

Question 3 :

In an AP

Given a = 5, d = 3, a n= 50 find n and Sn

Solution :

a_n = a + (n - 1) d

50 = 5 + (n - 1) (3)

50 - 5 = (n - 1) (3)

45 = (n - 1) (3)

45/3 = n - 1

 n - 1 = 15

 n = 15 + 1

 n = 16

S n = (n/2) [a + l]

S₁₆ = (16/2) [5 + 50]

  =  8 (55)

  =  440

Question 4 :

150 workers were engaed to finish a piece of work in certain number of days. Four workers dropped the second day, four more workers dropped the third day and so on. It takes 8 more days to finish the work now. Find the number of days in which the work was completed.

Solution :

Total number of workers = 150

150, 146, 142, ..........

By finding the 8th term, we find the number of workers in which the work was completed.

a = 150, d = 146 - 150 ==> -4 and n = 8

a₈ = 150 + (8 - 1) (-4)

= 150 + 7(-4)

= 150 - 28

= 122

So, there are 122 workers on that day.

Question 4 :

How many terms of the AP -15,-13,-11,… are needed to make the sum -55?

Solution :

Sn = -55

-15, -13, -11,…

a = -15, d = -13 + 15 ==> 2

(n/2)[2a + (n - 1) d] = -55

(n/2)[2a + (n - 1) d] = -55

(n/2) [2(-15) + (n - 1)2] = -55

(n/2) [(-15) + (n - 1)] = -55

n [n - 16] = -55

n² - 16n + 55 = 0

(n - 11)(n - 5) = 0

n = 11 and n = 5

Question 6 :

Solve the equation : (−4) + (−1) + 2 +⋯+ 𝑥 = 437

Solution :

a = -4, d = -1 + 4 ==> 3, l = x and S_n = 437

an = x

-4 + (n - 1) 3 = x

n - 1 = x + 4

n = x + 5

(n/2)[2a + (n - 1) d] = 437

(x + 5)/2 [2(-4) + (x + 4) (3)] = 437

(x + 5)/2 [-8 + 3x + 12] = 437

(x + 5) [3x + 4] = 437(2)

3x² + 4x + 15x + 20 = 874

3x² + 19x + 20 - 874 = 0

3x² + 19x - 854 = 0

3x² + 61x - 42x - 854 = 0

x(3x + 61) - 14(3x + 61) = 0 

(x - 14)(3x + 61) = 0

x = 14 and x = -61/3

Applying x = 14, we get n = 14 + 5 ==> 19

So, the value of x is 14.

Question 7 :

The sum of the first 7 terms of an AP is 63 and that of its next 7 terms is 161. Find the AP.

Solution :

S₇ = 63

(7/2)[2a + 6d] = 63

a + 3d = 63/7

a + 3d = 9 -----(1)

S₁₄ = 63 + 161

14/2 [2a + 13d] = 224

[2a + 13d] = 224/7

2a + 13d = 32  -----(2)

2(1) - (2)

2a + 6d - (2a + 13d) = 18 - 32

6d - 13d = -14

-7d = -14

d = 2

Applying the value of d, we get

a + 3(2) = 9

a + 6 = 9

a = 9 - 6

a = 3

3, 5, 7, .............. is the required arithmetic progression.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.