ARITHMETIC AND GEOMETRIC SEQUENCES WORKSHEET

1. Find the common difference and 15^th term of the arithmetic progression :

125, 120, 115, 110, ..........

2. Find the sum of the terms in the following arithmetic progression :

3 + 7 + 11 + 15 + ........ to 25 terms

3. Find the number of terms in the following arithmetic progression :

7, 12, 17, ........ 252

4. Find the common ratio and 15^th term of the geometric progression :

2, 4, 8, ..........

5. Find the sum of the terms in the following geometric progression :

2 + 6 + 18 + 54 + ........ to 10 terms

6. Divide 69 into three parts which are in arithmetic progression are such that the product of first two parts is 483.

7. If 5^th and 12^th terms of an arithmetic sequence are 14 and 35 respectively, find the arithmetic progression.

8. Find the sum : 

16 + 17 + 18 + .......... + 99

1. Answer :

125, 120, 115, 110, ..........

Common difference :

d = second term - first term

= 120 - 125

= -5

Formula to find n^th term of an arithmetic progression :

a_n = a₁ + (n - 1)d

Substitute n = 15, a₁ = 125 and d = -5.

a₁₅ = 125 + (15 - 1)(-5)

= 125 + (14)(-5)

= 125 - 70

= 55

2. Answer :

3 + 7 + 11 + 15 + ........ to 25 terms

a₁ = 3

d = second term - first term

= 7 - 3

= 4

Formula to find sum to 'n' terms of an arithmetic progression :

S_n = (n/2)[2a₁ + (n - 1)d]

Substitute n = 25, a₁ = 3 and d = 3.

S₂₅ = (25/2)[2(3) + (25 - 1)(4)]

= (25/2)[2(3) + (24)(4)]

= (25/2)[6 + 96]

= (25/2)[102]

= 25(51)

= 1275

3. Answer :

7, 12, 17, ........ 252 

a₁ = 7

 l = 252

d = second term - first term

= 12 - 7

= 5

Formula to find number of terms 'n' in the above arithmetic progression :

n = [(l - a)/d] + 1

Substitute a₁ = 7,  l = 252 and d = 5.

n = [(252 - 7)/5] + 1

= [(245)/5] + 1

= 49 + 1

= 50

4. Answer :

2, 4, 8, ..........

Common ratio :

r = second term/first term

= 4/2

= 2

Formula to find n^th term of a geometric progression :

a_r = a₁r^n-1

Substitute n = 15, a₁ = 2 and r = 2.

a₁₅ = 2(2)^{15 - 1}

= 2(2)¹⁴

= 2(16384)

= 32768

5. Answer :

2 + 6 + 18 + 54 + ........ to 10 terms

a₁ = 2

r = second term/first term

= 6/2

= 3 > 1

Formula to find sum to 'n' terms of a geometric progression when r > 1 :

S_n = a₁(rⁿ - 1)/( r - 1)

Substitute a₁ = 2, r = 3 and n = 10.

S₁₀ = 2(3¹⁰ - 1)/(3 - 1)

= 2(59049 - 1)/2

= 2(59048)/2

= 59048

6. Answer :

Since we divide 69 into three parts which are in arithmetic progression, the three parts of 69 can be assumed as

a - d, a, a + d

Then,

(a - d) + a + (a + d) = 69

a - d + a + a + d = 69

3a = 69

Divide both sides by 3.

a = 23

Given : The product of first two parts is 483.

(a - d)(a) = 483

Substitute a = 23.

(23 - d)(23) = 483

Divide both sides by 23.

23 - d = 21

Subtract 23 from both sides.

- d = -2

Multiply both sides by -1.

d = 2

The three parts of 69 which are in arithmetic progression are

(a - d), a, (a + d)

(23 - 2), 2, (23 + 2)

21, 2, 25

7. Answer :

(2) - (1) :

(a₁ + 11d) - (a₁ + 4d) = 35 - 14

a₁ + 11d - a₁ - 4d = 21

7d = 21

Divide both sides by 7.

d = 3

Substitute d = 3 in (1).

a₁ + 4(3) = 14

a₁ + 12 = 14

Subtract 12 from both sides.

a₁ = 2

Arithmetic progression :

a₁, a₁ + d, a₁ + 2d, a₁ + 3d, .........

Substitute a₁ = 2 and d = 3.

2, 2 + 3, 2 + 2(3), 2 + 3(3), .........

2, 2 + 3, 2 + 6, 2 + 9, .........

2, 5, 8, 11, .........

8. Answer :

Formula to find sum of first 'n' natural numbers :

1 + 2 + 3 + .......... + n = n(n + 1)/2

16 + 17 + 18 + .......... + 99 :

= (1 + 2 + 3 + .......... + 99) - (1 + 2 + 3 + .......... + 15)

= [99(99 + 1)/2] - [15(15 + 1)/2]

= [99(100)/2] - [15(16)/2]

= [99(50)] - [15(8)]

= 4950 - 120

= 4830

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