Find the eighteenth term of the A. P. : 1, 7, 13, 19, ..... (2 marks)

2. Find the eighteenth term of the A. P. : 1, 7, 13, 19, ..... (2 marks)

Sol. For the given A.P. 1, 7, 13, 19, .....

Here, a = t1 = 1

d = t2 – t1 = 7 – 1 = 6

We know,

tn = a + (n – 1) d

 t18 = a + (18 – 1) d

 t18 = 1 + 17 (6)

 t18 = 1 + 102

 t18 = 103

 Eighteenth term of A.P. is 103.

1, 4, 7, 10, .... (1 mark)

(iii) 1, 4, 7, 10, .... (1 mark)

Sol. t₁ = 1, t2 = 4, t3 = 7, t4 = 10

t₂ – t₁ = 4 – 1 = 3

t₃ – t₂ = 7 – 4 = 3

t₄ – t₃ = 10 – 7 = 3

Here, The difference between ANY two consecutive terms 3 which is constant.

∴ The sequence is an A.P.

3, 5, 7, 9, 11, ..... (1 mark)

(ii) 3, 5, 7, 9, 11, ..... (1 mark)

Sol. t₁ = 3, t2 = 5, t3 = 7, t4 = 9, t5 = 11

t₂ – t₁ = 5 – 3 = 2

t₃ – t₂ = 7 – 5 = 2

t₄ – t₃ = 9 – 7 = 2

t₅ – t₄ = 11 – 9 = 2

Here, The difference between ANY two consecutive terms is 2 which is constant.

∴ The sequence is an A.P.

(iii) 1, 3, 7, 15, 31, .... (1 mark)

(iii) 1, 3, 7, 15, 31, .... (1 mark)

Sol. t₁ = 1

t₂ = t₁ + 2 = 1+ 2 = 3

t₃ = t₂ + 4 = 3 + 4 = 7

t₄ = t₃ + 8 = 7 + 8 = 15

t₅ = t₄ + 16  = 15 + 16 = 31

t₆ = t₅ + 32  = 31 + 32 = 63

t₇ = t₆  + 64  = 63 + 64 = 127

t₈ = t₇  + 128  = 127 + 128 = 255

t₉ = t₈  + 256  = 255 + 256 = 511

∴ The next four terms of the sequence are 63, 127, 255 and 511.

For each sequence, find the next four terms. (i) 1, 2, 4, 7, 11, ..... (1 mark)

1. For each sequence, find the next four terms.

(i) 1, 2, 4, 7, 11, ..... (1 mark)

Sol. t₁ = 1 + 0 = 1

t₂ = 1 + 1 = 2

t₃ = 2 + 2 = 4

t₄ = 4 + 3 = 7

t₅ = 7 + 4 = 11

t₆ = 11 + 5 = 16

t₇ = 16 + 6 = 22

t₈ = 22 + 7 = 29

t₉ = 29 + 8 = 37

∴ The next four terms of the sequence are 16, 22, 29 and 37.