Arithmetic Sequence Recursive Formula & Solved Examples

Ans. 

1. The difference between 3 and 1 is 2, 5 and 3 is 2, 7 and 5 is 2

Thus the common difference ‘d’ is 2 for the given sequence

2. The difference between 4 and 9 is 5, 14 and 9 is 5, 19 and 14 is 5

Thus the common difference ‘d’ is 5 for the given sequence

100. The difference between 11 and 6 is 5, 16 and 11 is 5, 21 and 16 is 5

Thus the common difference ‘d’ is 5 for the given sequence

500. The difference between 4.5 and 3 is 1.5, 6 and 4.5 is 1.5, 7.5 and 6 is 1.5

Thus the common difference ‘d’ is 1.5 for the given sequence

Ans.

1. a_n = a_n−1 + d

a₃ = a₃₋₁ + d

a₃ = a₂ + d

a₃ = 21 + 4

a3 = 25

2. a_n = a_n−1 + d

a₂ = a₂₋₁ + d

a₂ = a₁ + d

a₂ = 50 + 2

a₂ = 52

100. a_n = a_n−1 + d

a₃ = a₃₋₁ + d

a₃ = a₂ + d

a₃ = 12 + 4

a₃ = 16

500. a_n = a_n−1 + d

a₈ = a₈₋₁ + d

a₈ = a₇ + d

a₈ = 32 + 4

a₈ = 36

Ans. The common difference of the arithmetic sequence 1, 5/4, 3/2, 7/4 … is

d = 5/4 – 1, 5/4 – 3/2, = 1/4

thus, recursive formula –

a_n = a_n−1 + d

a_n = a_n−1 + 1/4

For finding the 5th term

a₅ = a₅₋₁ + 1/4

a₅ = a⁴ + 1/4

a₅ = 7/4 + 1/4

a₅ = 2/1

a₆ = a₆₋₁ + 1/4

a₆ = a₅ + 1/4

a₆ = 2/1 + 1/4

a₆ = 9/4

The next two terms in the sequence is 2/1, 9/4.

Ans. 

1. Here a_n = 2n + 5

Substituting n =1, 2, 3, in the above equation we get

a₁ =2(1) + 5 = 7

a₂ = 9 

a₃ = 11

Therefore, the required terms are 7, 9, and 11.

2. Here a_n = \(\frac{n-3}{4}\)

Thus, 

a₁ = \(\frac{1-3}{4}\) = -\(\frac{1}{2}\)

a₂ =\(\frac{2-3}{4}\) = -\(\frac{1}{4}\)

a₃ = \(\frac{3-3}{4}\)= 0

Therefore, the first three terms are -\(\frac{1}{2}\), -\(\frac{1}{4}\), and 0.

Ans. Putting n = 20, we obtain 

a₂₀ = (20 – 1) (2 – 20) (3 + 20)

= 19 × (– 18) × (23) 

= – 7866.

Putting n = 25, we obtain

a₂₅ = (25 – 1) (2 – 25) (3 + 25)

= 24 × (– 23) × (28) 

= – 15,456.

Thus the 20^th and 25^th terms of the sequence – 7866 and – 15,456.

Ans. Let’s consider,

a₁ = 1, 

a₂ = a₁ + 2 = 1 + 2 = 3, 

a₃ = a₂ + 2 = 3 + 2 = 5,

a₄ = a₃ + 2 = 5 + 2 = 7, 

a₅ = a₄ + 2 = 7 + 2 = 9.

Hence, the first five terms of the sequence are 1,3,5,7, and 9. 

The corresponding series is 1 + 3 + 5 + 7 + 9 +...

Ans. The first term of the given arithmetic sequence is, a1 = −1

The given recursive formula is,

a_n = a_n−1 − 3

Substituting n = 2 in the given formula, 

a₂ = a₁ − 3 

= −1 − 3 

= − 4.

Substituting n = 3 in the given formula, 

a₃ = a₂ − 3 

= −4 − 3 

= − 7

Substitute n = 4 in the given formula, 

a₄ = a₃ − 3 

= −7 − 3 

= − 10

Substitute n = 5 in the given formula, 

a₅ = a₄ − 3 

= −10 − 3 

= − 13

Substitute n = 6 in the given formula, 

a₆ = a₅ − 3 

= −13 − 3 

= − 16

The first 6 terms of the given arithmetic sequence are, -1, -4, -7, -10, -13 and -16.

Ans. Here, 

d = 9 – 3 = 6

n = x is in the 4 term of the sequence

a_n = a_n−1 + d

a₄ = a₄₋₁ + 6

a₄ = a₃ + 6

a₄ = 15 + 6 … (a3 is given in the sequence as 15)

a₄ = 21

thus the value of the fourth term of the arithmetic sequence ‘x’ is 21.

Ans. Using arithmetic sequence recursive formula,

a_n = a_n−1 + d

d = 7 – 5 

d = 2

As 5 terms of the sequence are already found, thus we will find 6^th, 7^th, and 8^th terms 

a₆ = a₆₋₁ + 2

a₆ = a₅ + 2

a₆ = 19 + 2 … (a5 is given in the sequence as 19)

a₆ = 21 … (1)

a₇ = a₇₋₁ + 2

a₇ = a₆ + 2

a₇ = 21 + 2 … (from 1)

a₇ = 23 … (2)

a₈ = a₈₋₁ + 2

a₈ = a₇ + 2

a₈ = 23 + 2 … (from 2)

a₈ = 25

Thus the 3 consecutive terms are 21, 23, and 25.

Ans.

a₁ = – 4

And

a_n = a_n−1 + (-2)

thus d = - 2

next step, n = 2

a₂= a₂₋₁ + (-2)

a₂= a₁ + (-2)

a₂ = - 4 - 2

a₂ = - 6

n = 3

a₃ = a₃₋₁ + (-2)

a₃= a₂ + (-2)

a₃ = - 6 - 2

a₃ = - 8

n = 4

a₄ = a₄₋₁ + (-2)

a₄ = a₃ + (-2)

a₄ = - 8 - 2

a₄ = - 10

Thus the first four terms of the sequence are - 4, - 6, - 8, - 10…