FINDING THE INDICATED TERM OF A SEQUENCE

Question 1 :

Find the indicated terms of the sequences whose nth terms are given by

(i) an  =  5n / (n + 2); a6 and a13

Solution :

To find a6, we have to apply 6 instead of n.

To find a13, we have to apply 13 instead of n.

 an  =  5n / (n + 2)

n = 6

 a6  =  5(6)/(6 + 2)

  =  30/8

 a6  =  15/4

 an  =  5n / (n + 2)

n = 13

 a13  =  5(13)/(13 + 2)

  =  65/15

 a13  =  13/3

(ii) a =  -(n2 - 4); a4 and a11

Solution :

To find a4, we have to apply 4 instead of n.

To find a11, we have to apply 11 instead of n.

 an  =   -(n2 - 4)

n = 4

 a4  =   -(42 - 4)

 a4  =   -(16 - 4)

=  -12

 an  =   -(n2 - 4)

n = 11

 a11  =   -(112 - 4)

 a11  =   -(121 - 4)

=  -117

Question 2 :

Find a8 and a15 whose nth term is

Solution :

In a8, 8 is even in a15, 15 is odd 

an  =  (n2 - 1)/(n + 3)

a8  =  (82 - 1)/(8 + 3)

  =  (64 - 1)/11

  =  63/11

an  =  n2/(2n + 1)

a15  =  152/(2(15) + 1)

  =  225/(30 + 1)

  =  225/31

Question 3 :

If a1 = 1, a2 = 1 and an = 2an - 1 + an - 2≥ 3, n ∈ N, then find the first six terms of the sequence.

Solution :

First and second terms are 1.

an = 2an - 1 + an - 2 

n = 3

an = 2an - 1 + an - 2 

a3 = 2a3-1 + a3-2 

a3  =  2a2 + a1

a3  =  2(1) + 1

a3  =  3

n = 4

an = 2an - 1 + an - 2 

a4 = 2a4-1 + a4-2 

a4  =  2a3 + a2

a4  =  2(3) + 1

a4  =  7

n = 5

an = 2an - 1 + an - 2 

a5 = 2a5-1 + a5-2 

a5  =  2a4 + a3

a5  =  2(7) + 3

a5  =  17

n = 6

an = 2an - 1 + an - 2 

a6 = 2a6-1 + a6-2 

a6  =  2a5 + a4

a5  =  2(17) + 7

a5  =  41

Hence the first 6 terms are 1, 1, 3, 7, 17, 41.

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