FINDING nth TERM OF A SEQUENCE

There is some arrangement or pattern followed in every sequence. We find nth term of the sequence in term of n. By applying the position value of n in the nth term, we must get the value in that particular place.

For example, by applying 5 instead of n in nth term, we get the 5th term of the sequence.

Question 1 :

Find the nth term of the following sequences

(i) 2, 5, 10, 17,...

Solution :

(1 + 1), (4 + 1), (9 + 1), (16 + 1),...

(12 + 1), (22 + 1), (32 + 1), (42 + 1),...

nth term of the sequence  :

an  =  n2 + 1

(ii) 0, 1/2, 2/3, ............

Solution :

Denominator is 1 more than the numerator. By approaching the above problem in this way, we will not get the first term.

Numerator is 1 less than the denominator.

  =  0, 1/2, 2/3, ............

  =  (1-1)/1, (2-1)/2, (3-1)/3, ...................

nth term of the sequence is (n - 1)/n

(iii) 3, 8, 13, 18,...

Solution :

  =  3, 8, 13, 18,...

we may see the common things followed in each term.

  =  (5 - 2), (10 - 2), (15 - 2), (20 - 2),...

nth term of the sequence is 5n - 2.

Finding the Indicated Term of a Sequence ?

Question 2 :

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

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