Examples

A: Find the rule for the following sequences:

Ex 1:

1, 3, 5, 7, 9, . . . .

Sol:

Consider the first sequence 1, 3, 5, 7, 9 . . . .
First term = t1 = 1 = 2 * 1 – 1
Second term = t2 = 3 = 2 * 2 – 1
Third term = t3 = 5 = 2 * 3 – 1
Fourth term = t4 = 7 = 2 * 4 – 1
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From this it is known that: tn = nth term = 2n – 1
∴ The rule for the sequence 1, 3, 5, 7, . . . . is 2n – 1 where n ∈ N
Ex 2:

5, 8, 11, 14, 17 . . . .

Sol:

Consider the second sequence 5, 8, 11, 14, 17 . . . .
First term = t1 = 5 = 3 * 1 + 2
Second term = t2 = 8 = 3 * 2 + 2
Third term = t3 = 11 = 3 * 3 + 2
Fourth term = t4 = 14 = 3 * 4 + 2
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Hence tn = nth term = 3n + 2
∴ The rule for the sequence 5, 8, 11, 14, . . . . is 3n + 2 where n ∈ N.
Ex 3:

0, – 1, – 2, – 3, – 4. . . .

Sol:

Consider the third sequence 0, – 1, – 2, – 3, – 4. . . .
First term = t1 = 0 = 1 – 1
Second term = t2 = – 1 = 1 – 2
Third term = t3 = – 2 = 1 – 3
Fourth term = t4 = – 3 = 1 – 4
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Hence tn = nth term = 1 – n
∴ The rule for the sequence 0, – 1, – 2, – 3, . . . . is 1 – n where n ∈ N.

B: Write the first four terms in each of the sequence defined by:

Ex 1:

tn = n(n – 1)

Sol:

Given, tn = n(n – 1)
Putting n = 1, 2, 3, 4 we get first four terms.
Thus,
t1 = 1(1 – 1) = 1(0) = 0
t2 = 2(2– 1) = 2(1) = 2
t3 = 3(3– 1) = 3(2) = 6
t4 = 4(4– 1) = 4(3) = 12
∴ First four terms of the given sequence are 0, 2, 6, 12
tn = n2 + 1
Ex 2:

tn = (n2 + 1)

Sol:

Given, tn = (n2 + 1)
Putting n = 1, 2, 3, 4 we get first four terms.
Thus,
t1 = 12 + 1 = 1 + 1 = 2
t2 = 22 + 1 = 4 + 1 = 5
t3 = 32 + 1 = 9 + 1 = 10
t4 = 42 + 1 = 16 + 1 = 17
∴ First four terms of the given sequence are 2, 5, 10, 17
tn = n3 + n2 + n + 1.
Ex 3:

tn = n3 + n2 + n + 1

Sol:

Given, tn = n3 + n2 + n + 1 Putting n = 1, 2, 3, 4 we get first four terms.
Thus,
t1 = 13 + 12 + 1 + 1 = 1 + 1 + 1 + 1 = 4
t2 = 23 + 22 + 2 + 1 = 8 + 4 + 2 + 1 = 15
t3 = 33 + 32 + 3 + 1 = 27 + 9 + 3 + 1 = 40
t4 = 43 + 42 + 4 + 1 = 64 + 16 + 4 + 1 = 85
∴ First four terms of the given sequence are 4, 15, 40, 85

C: Find the terms t12, t20 in the sequence whose nth term is .

Sol:

Given tn =

By substituting n = 12, 20 we get the terms t12, t20