FIND THE SUM OF ARITHMETIC SERIES

Find the sum of

Problem 1 :

11 + 14 + 17 + … to 16 terms

Solution :

11 + 14 + 17 + … to 16 terms

a = 11, d = 14 – 11 ==> 3 and n = 16 

S_n = (n/2)(2a + (n – 1)d)

= 16/2(2(11) + (16 – 1)3)

= 8(22 + (15)3)

= 8(22 + 45)

= 8(67)

S₁₆ = 536

Problem 2 :

27 + 22 + 17 + … to 10 terms

Solution :

27 + 22 + 17 + … to 10 terms

a = 27, d = 22 – 27 ==> -5 and n = 10

S_n = (n/2)(2a + (n – 1)d)

= 10/2(2(27) + (10 – 1)(-5))

= 5(54 + (9 × (-5))

= 5(54 - 45)

= 5(9)

S₁₀ = 45

Problem 3 :

5 + 17 + 29 + … + 161

Solution :

5 + 17 + 29 + … + 161

a = 5, d = 17 – 5 ==> 12

In the series above, we have last term. To find the number of terms, we use the formula

n = [(161 - 5)/12] + 1

n = (156/12) + 1

n = 13 + 1

n = 14

So, number of terms in the series is 14.

S_n= (n/2)(2a + (n – 1)d)

S₁₄ = (14/2)(2(5) + (14 – 1)12)

= 7(10 + (13)(12))

= 7(10 + 156)

= 5(166)

S₁₄ = 830

Problem 4 :

7.2 + 7.8 + 8.4 + … to 21terms

Solution :

7.2 + 7.8 + 8.4 + … to 21terms

a = 7.2, d = 7.8 – 7.2 ==> 6 and n = 21

S_n= (n/2) (2a + (n – 1)d)

S₂₁ = (21/2) (2(7.2) + (21 – 1)6)

= 10.5 (14.4 + (20)6)

= 10.5 (14.4 + 120)

= 10.5 (134.4)

S₂₁ = 1411.2

Problem 5 :

90 + 79 + 68 + … -20

Solution :

90 + 79 + 68 + … -20

a = 90, d = 79 – 90 ==> -11

In the series above, we have last term. To find the number of terms, we use the formula

n = [(-20 - 90)/(-11)] + 1

n = (-110/(-11)) + 1

n = 10 + 1

 = 11

S_n = (n/2) (2a + (n – 1)d)

= (11/2) (2(90) + (11 – 1)(-11))

= 5.5 (180 + 10(-11))

= 5.5 (180 – 110)

= 5.5 (70)

S₁₁ = 385

Problem 6 :

0.12 + 0.155 + 0.19 + … to 150 terms

Solution :

0.12 + 0.155 + 0.19 + … to 150 terms

a = 0.12, d = 0.155 – 0.12 ==> 0.035

n = 150

S_n = (n/2) (2a + (n – 1)d)

S₁₅₀ = (150/2) (2(0.12) + (150 – 1)0.035)

= 75 (0.24 + (14)0.035)

= 75(0.24 + 0.49)

= 75(0.73)

S₁₅₀ = 54.75