ARITHMETIC AND GEOMETRIC SERIES WORKSHEET

Problem 1 :

Evaluate the following arithmetic series :

7 + 11 + 15 + 19 + ............ to 56 terms

Problem 2 :

Evaluate the following arithmetic series :

3 + 6 + 9 + 12 + ............ + 252

Problem 3 :

Evaluate the following geometric series :

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

Problem 4 :

Evaluate the following geometric series :

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

Problem 5 :

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

7 + 7 + 7 + ............

Problem 6 :

Find the sum of the following infinite geometric series :

2 + 1 + 1/2 +.......... ∞

Problem 7 :

Find the sum of the first 20-terms of the arithmetic progression having the sum of first 10 terms as 52 and the sum of the first 15 terms as 77.

Problem 8 :

Find the sum up to 17^th term of the following series :

Answers

1. Answer :

7 + 11 + 15 + 19 + ............ to 56 terms

Formula to find the sum of first n terms of an arithmetic series (when number of terms is known) :

S_n = (n/2){2t₁ + (n - 1)d}

From the given arithmetic series, we have

t₁ = 7

d = t₂ - t₁

= 11 - 7

= 4

n = 56

Substitute n = 56, t₁ = 7 and d = 4 into the formula given above.

S₅₆ = (56/2){2(7) + (56 - 1)(4)}

= 28{14 + (55)(4)}

= 28{14 + 220}

= 28{234}

= 6552

2. Answer :

3 + 6 + 9 + 12 + ............ + 252

Formula to find the sum of first n terms of an arithmetic series (when last term l  is known) :

S_n = (n/2){a + l}

From the given arithmetic series, we have

t₁ = 3

d = t₂ - t₁

= 6 - 3

= 3

n = [(l - t₁)/d] + 1

= [(252 - 3)/3] + 1

= [249/3] + 1

= 83 + 1

= 84

Substitute n = 84, t₁ = 3 and l = 252 into the formula given above.

S₈₄ = (84/2){3 + 252}

= 42(255)

= 10,710

3. Answer :

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

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

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

4. Answer :

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

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

5. Answer :

7 + 7 + 7 + ............

From the given geometric series, we have

t₁ = 7

r = t₂/t₁

= 7/7

= 1

n = n

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

S_n = nt₁

Substitute t₁ = 7.

S_n = n(5)

S_n = 7n

6. Answer :

2 + 1 + 1/2 +.......... ∞

t₁ = 2

r = t₂/t₁ = 1/2

The value of r (= 1/2) 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_∞ = 2/(1 - 1/2)

= 2/(1/2)

= 2(2/1)

= 4

7. Answer :

Formula to find the sum of first n terms of an arithmetic series (when number of terms is known) :

S_n = (n/2){2t₁ + (n - 1)d}

Given : Sum of first 10 terms is 52.

S₁₀ = 52

(10/2){2t₁ + (10 - 1)d} = 52

5{2t₁ + 9d} = 52

10t₁ + 45d = 52 ----(1)

Given : Sum of first 15 terms is 77.

S₁₅ = 52

(15/2){2t₁ + (15 - 1)d} = 77

(15/2){2t₁ + 14d} = 77

(15/2) ⋅ 2{t₁ + 7d} = 77

15{t₁ + 7d} = 77

15t₁ + 105d = 77 ----(2)

3 ⋅ (1) - 2 ⋅ (2) :

-75d = 2

d = -2/75

d = -2/75 in (1).

10t₁ + 45(-2/75) = 52

10t₁ + 3(-2/5) = 52

10t₁ - 6/5 = 52

Add 6/5 from both sides.

10t₁ = 52 + 6/5

10t₁ = (260 + 6)/5

10t₁ = 266/5

10t₁ = 254/5Divide both sides by 10 and simplify.

t₁ = 133/25

Sum of first 20 terms :

8. Answer :

First let us find the n^th term of the series.

In numerator, we have sum of cubes of first n natural numbers and in denominator, we have sum of first odd natural numbers.

Let t_n be the n^th term of the series.

Take summation on both sides.

Substitute n = 17.

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