FIND THE SUM OF N TERMS OF ARITHMETIC SERIES

Find the indicated nth partial sum (S_n) of the arithmetic sequence.

Problem 1 :

8, 20, 32, 44, … n = 10

Solution :

8, 20, 32, 44, … n = 10

a = 8

 d = a₂ – a₁ ==> 20 – 8

d = 12

s_n= (n/2)(2a + (n - 1) d)

s₁₀= (10/2)(2(8) + (10 - 1) 12)

s₁₀= 5(16 + 108)

s₁₀= 5(124)

s₁₀= 620

Problem 2 :

a₁ = -6, d = 4, n = 50

Solution :

a₁ = -6, d = 4, n = 50

a_n = a₁ + (n – 1)d

a₅₀ = -6 + (50 – 1)4

= -6 + (49) × 4

= -6 + 196

= 190

S_n = n/2(a₁ + a_n)

S₅₀ = 50/2(a₁ + a₅₀)

= 25(-6 + 190)

= 25(184)

S₅₀ = 4600

Problem 3 :

100 + 105 + 110 + … + 220

Solution :

100 + 105 + 110 + … + 220

a₁ = 100, a₂ = 105

d = a₂ – a₁

= 105 – 100

d = 5

t_n = a₁ + (n – 1)d

220 = 100 + (n – 1)5

Subtracting 100 on both sides.

120 = 5n – 5

120 + 5 = 5n

125 = 5n

Dividing 5 on both sides.

125/5 = 5n/5

25 = n

Since we know the last term, we can use the formula

S_n = (n/2)(a₁ + a_n)

S₂₅ = (25/2)(100 + 220)

S₂₅ = (25/2)(320)

S₂₅ = 25(160)

S₂₅ = 4000

Problem 4 :

0.5 + 1.3 + 2.1 + … + 70.1

Solution :

0.5 + 1.3 + 2.1 + … + 70.1

a₁ = 0.5

a₂ = 1.3

d = a₂ – a₁

= 1.3 – 0.5

d = 0.8

t_n = a₁ + (n – 1)d

70.1 = 0.5 + (n – 1)0.8

69.6 = 0.8n – 0.8

Adding 0.8 on both sides.

69.6 + 0.8 = 0.8n

70.4 = 0.8n

88 = n

S_n = n/2(a₁ + a_n)

S₈₈ = (88/2) (a₁ + a₈₈)

= 44(0.5 + 70.1)

= 44(70.6)

S₈₈ = 3106.4

Problem 5 :

a₁ = 3, d = 2, n = 12

Solution :

a₁ = 3, d = 2, n = 12

a_n = a₁ + (n – 1)d

a₁₂ = 3 + (12 – 1)2

= 3 + (11) × 2

= 3 + 22

= 25

S_n = n/2(a₁ + a_n)

S₁₂ = 12/2(a₁ + a₁₂)

= 6(3 + 25)

= 6(28)

S₁₂ = 168

Problem 6 :

a₁ = 100, d = -5, n = 8

Solution :

a₁ = 100, d = -5, n = 8

a_n = a₁ + (n – 1)d

a₈ = 100 + (8 – 1)(-5)

= 100 + (7) × (-5)

= 100 - 35

= 65

S_n = n/2(a₁ + a_n)

S₈ = 8/2(a₁ + a₈)

= 4(100 + 65)

= 4(165)

S₈ = 660

Problem 7 :

a₂ = 8, a₅ = 9.5, n = 12

Solution :

a₂ = 8, a₅ = 9.5, n = 12

a_n = a₁ + (n – 1)d

a₂ = a₁ + (2 – 1)d

8 = a₁ + 1d --- (1)

a₅ = a₁ + (5 – 1)d

9.5 = a₁ + 4d --- (2)

From (1) and (2)

-1.5 = -3d

Dividing -3 on both sides.

-1.5/-3 = -3/-3d

0.5 = d

Substitute (0.5) in equation (1)

8 = a₁ + 1(0.5)

8 = a₁ + 0.5

Subtracting 0.5 on both sides.

8 – 0.5 = a₁ + 0.5 – 0.5

7.5 = a₁

a_n = a₁ + (n – 1)d

a₁₂ = 7.5 + (12 – 1)0.5

= 7.5 + (11)0.5

= 7.5 + 5.5

a₁₂ = 13

S_n = n/2(a₁ + a_n)

S₁₂ = 12/2(7.5 + a₁₂)

= 6(7.5 + 13)

= 6(20.5)

S₁₂ = 123

Problem 8 :

-3 + (-3/2) + 0 + … + 30

Solution :

-3 + (-3/2) + 0 + … + 30

a₁ = -3

a₂ = -3/2

d = -3/2 + 3

= -3/2 + 3/1 × 2/2

= -3/2 + 6/3

d = 3/2

t_n = a₁ + (n – 1)d

30 = -3 + (n – 1)3/2

Adding 3 on both sides.

30 + 3 = (n – 1)3/2

33 = 3/2n – 3/2

Adding 3/2 on both sides.

33 + 3/2 = 3/2n

66/2 + 3/2 = 3/2n

69/2 = 3/2n

23 = n

S_n = (n/2)(a₁ + a_n)

S₂₃ = (23/2)(a₁ + a₂₃)

= (23/2)(-3 + 30)

= (23/2) (27)

S₂₃ = 621/2