0.5, 2, 3.5, 5, ...

(vi) 0.5, 2, 3.5, 5, ... (1 mark)

Sol. t1 = 0.5, t2 = 2, t3 = 3.5, t4 = 5

t2 – t1 = 2 – 0.5 = 1.5

t3 – t2 = 3.5 – 2 = 1.5

t4 – t3 = 5 – 3.5 = 1.5

Here, The difference between ANY two consecutive terms is 1.5 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.

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

(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.

Mohan has a hobby of solving problems every day. He solved 2, 4, 3, 2, 4, 5, 1 problems in the last week. What is the mean number of problems solved by Mohan per day?

Solution:

Since Mohan solved 2, 4, 3, 2, 4, 5, 1 problems in the last week,

∴ Mean = Σx = 2 + 4 + 3 + 2 + 4 + 5 + 1 = 21 = 3 n 7 7

∴ Mean = 3

It means on an average Mohan solved 3 problems per day.