GEOMETRIC SERIES PROBLEMS AND SOLUTIONS

Problem 1 :

Evaluate the following geometric series :

2 + 4 + 8 + 16 + ............ to 15 terms

Solution :

Formula to find the sum of first n terms of a geometric :

S_n = [t₁(rⁿ - 1)]/(r - 1)

From the given geometric series, we have

t₁ = 2

r = t₂/t₁

= 6/3

= 2

n = 15

Substitute n = 15, t₁ = 2 and r = 2 into the formula given above.

S₁₅ = [2(2¹⁵ - 1)]/(2 - 1)

= [2(32768 - 1)]/1

= 2(32767)

= 65534

Problem 2 :

Evaluate the following geometric series :

2 + 6 + 12 + 24 + ............ + 354294

Solution :

Formula to find the sum of first n terms of a geometric series :

S_n = [t₁(rⁿ - 1)]/(r - 1)

From the given geometric series, we have

t₁ = 2

r = t₂/t₁

= 6/3

= 2

Let t_n = 354294.

t_n = 354294

t₁r^n-1 = 354294

2(3)^n-1 =354294

Divide both sides by 2.

3^{n - 1} = 177147

3^{n - 1} = 3¹¹

n - 1 = 11

n = 12

Substitute n = 12, t₁ = 2 and r = 3 into the formula given above.

S₁₂ = [2(3¹² - 1)]/(3 - 1)

= [2(531441 - 1)]/2

= [2(531440)]/2

= 531440

Problem 3 :

Find the sum of n terms of the following geometric series.

5 + 5 + 5 + ............

Solution :

From the given geometric series, we have

t₁ = 5

r = t₂/t₁

= 5/5

= 1

n = n

Formula to find the sum of first n terms of a geometric series (when r = 1) :

S_n = nt₁

Substitute t₁ = 5.

S_n = n(5)

S_n = 5n

Problem 4 :

Find the sum of the following infinite geometric series :

3 + 1 + 1/3 +.......... ∞

Solution :

t₁ = 3

r = t₂/t₁ = 1/3

The value of r (= 1/3) is in the interval -1 < r < 1.

So, the sum for the given infinite geometric series exists.

Formula to find the sum of infinite geometric series :

S_∞ = t₁/(1 - r)

Substitute t₁ = 3 and r = 1/3.

S_∞ = 3/(1 - 1/3)

= 3/(2/3)

= 3(3/2)

= 9/2

Problem 5 :

Compute the sum of first n terms of the following series :

8 + 88 + 888 + ............

Solution :

This series is not a geometric series. However, it can be changed to geometric series as shown below.

= 8 + 88 + 888 + ............ to n terms

= 8(1 + 11 + 111 + ............ to n terms)

= 8 ⋅ (9/9)(1 + 11 + 111 + ............ to n terms)

  = (8/9)(9 + 99 + 999 + ............ to n terms)

= (8/9)[(10 - 1) + (100 - 1) + (1000 - 1) + ............ to n terms]

= (8/9)[(10 + 100 + 1000 + ............ to n terms) - (1 + 1 + 1 + .......... to n terms)]

Problem 6 :

Compute the sum of first n terms of the following series :

6 + 66 + 666 + ............

Solution :

This series is not a geometric series. However, it can be changed to geometric series as shown below.

= 6 + 66 + 666 + ............

= 6(1 + 11 + 111 + ............ to n terms)

= 6 ⋅ (9/9)(1 + 11 + 111 + ............ to n terms)

  = (6/9)(9 + 99 + 999 + ............ to n terms)

= (2/3)[(10 - 1) + (100 - 1) + (1000 - 1) + ............ to n terms]

= (2/3)[(10 + 100 + 1000 + ............ to n terms) - (1 + 1 + 1 + .......... to n terms)]

Problem 7 :
Compute the sum of first n terms of the following series :

1 + (1 + 4) + (1 + 4 + 4²) + (1 + 4 + 4² + 4³) + ............

Solution :

t₁ = 1

t₂ = 1 + 4

t₃ = 1 + 4 + 4²

t₄ = 1 + 4 + 4² + 4³

t_n = 1 + 4 + 4² + 4³ + ............ to n terms

Take summation on both sides.

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