Form two digit numbers using the digits 0, 1, 2, 3, 4, 5 without repeating the digits. P is the event that the number so formed is even. Q is the event that the number so formed is divisible by 3. R is the event that the number so formed is greater than 50.

(v) Form two digit numbers using the digits 0, 1, 2, 3, 4, 5 without repeating the digits.

P is the event that the number so formed is even.

Q is the event that the number so formed is divisible by 3.

R is the event that the number so formed is greater than 50.

Sol. The two digit numbers that can be formed using the digits 0, 1, 2, 3, 4, 5 without repeating digits are as follows :

S = { 10, 12, 13, 14, 15, 20, 21, 23, 24, 25, 30, 31, 32, 34, 35, 40, 41, 42, 43, 45, 50, 51, 52, 53, 54 }

n (S) = 25

P is the event that number so formed is even

P = { 10, 12, 14, 20, 24, 30, 32, 34, 40, 42, 50, 52, 54 }

∴  n (P) = 13

Q is the event that number so formed is divisible by 3

Q = { 12, 15, 21, 24, 30, 42, 45, 51, 54 }

∴  n (Q) = 9

R is the event that the number so formed is greater than 50

R = { 51, 52, 53, 54 }

∴  n (R) = 4

4, 3, 2, 1, ....

(vii) 4, 3, 2, 1, .... (1 mark)

Sol. t1 = 4, t2 = 3, t3 = 2, t4 = 1,

t2 – t1 = 3 – 4 = – 1

t3 – t2 = 2 – 3 = – 1

t4 – t3 = 1 – 2 = – 1

Here, The difference between ANY two consecutive terms is –1 which is constant.

∴ The sequence is an A.P.

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.

(ix) 2, 4, 8, 16, ..... (1 mark)

(ix) 2, 4, 8, 16, ..... (1 mark)

Sol. t₁ = 2¹ = 2

t₂ = 2² = 4

t₃ = 2³ = 8

t₄ = 2⁴ = 16

t₅ = 2⁵ = 32

t₆ = 2⁶ = 64

t₇ = 2⁷ = 128

t₈ = 2⁸ = 256

∴  The next four terms of the sequence are 32, 64, 128 and 256.

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.

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.

Find the next two terms of the given sequence 2, 4, 8, 16, ...........