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.
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) an = -(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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