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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.

(vii) 2, 5, 8, 11, .... (1 mark)

Sol. t₁ = 2

t₂ = 2 + 3 = 5

t₃ = 5 + 3 = 8

t₄ = 8 + 3 = 11

t₅ = 11 + 3 = 14

t₆ = 14 + 3 = 17

t₇ = 17 + 3 = 20

t₈ = 20 + 3 = 23

∴ The next four terms of the sequence are 14, 17, 20 and 23.

(vi) 0.1, 0.01, 0.001, 0.0001, .... (1 mark)

Sol. t₁ = 0.1

t₂ = 0.1 × 0.1 = 0.01

t₃ = 0.01 × 0.1 = 0.001

t₄ = 0.001 × 0.1 = 0.0001

t₅ = 0.0001 × 0.1 = 0.00001

t₆ = 0.00001 × 0.1 = 0.000001

t₇ = 0.000001 × 0.1 = 0.0000001

t₈ = 0.0000001 × 0.1 = 0.00000001

The next four terms of the sequence are 0.00001, 0.000001,

0.0000001 and 0.00000001.

(iv) 192, – 96, 48, – 24, .... (1 mark)

Sol. t₁ = 192

t₂ = ¹⁹² / _–₂ = – 96

t₃ = ^–⁹⁶ / _–₂ = 48

t₄ = ⁴⁸ / _–₂ = – 24

t₅ = ^–²⁴ / _–₂ = 12

t₆ = ¹² / _–₂ = – 6

t₇ = ^–⁶ / _–₂ = 3

t₈=³/ _–₂ = ³/ _–₂

∴ The next four terms of the sequence are 12, – 6, 3 and ³/_–₂

(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.

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

(vii) 2, 5, 8, 11, .... (1 mark)

(vi) 0.1, 0.01, 0.001, 0.0001, .... (1 mark)

(iv) 192, – 96, 48, – 24, .... (1 mark)

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