Write the first 6 terms of the sequences whose nth term an is given below.
(i) an = n + 1 if n is odd
n if n is even
Solution :
If n = 1 an = n + 1 = 1 + 1 = 2 |
If n = 3 an = n + 1 = 3 + 1 = 4 |
If n = 5 an = n + 1 = 5 + 1 = 6 |
If n = 2 an = n a2 = 2 |
If n = 4 an = n a4 = 4 |
If n = 6 an = n a6 = 6 |
Hence first 6 terms are 2, 2, 4, 4, 6, 6
Question 2 :
an = 1 if n = 1
2 if n = 2
an−1 + an−2 if n > 2
Solution :
1st term = 1
2nd term = 2
3rd term :
an = an−1 + an−2
a3 = a3−1 + a3−2 ==> a2 + a1 ==> 2 + 1 ==> 3
4th term :
an = an−1 + an−2
a4 = a4−1 + a4−2 ==> a3 + a2 ==> 3 + 2 ==> 5
5th term :
an = an−1 + an−2
a5 = a5−1 + a5−2 ==> a4 + a3 ==> 5 + 3 ==> 8
6th term :
an = an−1 + an−2
a6 = a6−1 + a6−2 ==> a5 + a4 ==> 8 + 5 ==> 13
Hence the first 6 terms are 1, 2, 3, 5, 8 and 13.
Question 3 :
an = n if n is 1, 2 or 3
an−1 + an−2 + an−3 if n > 3
Solution :
1st term : an = n a1 = 1 |
2nd term : an = n a2 = 2 |
3rd term : an = n a3 = 3 |
4th term :
an = an−1 + an−2 + an−3
a4 = a3 + a2 + a1 ==> 3 + 2 + 1 ==> 6
5th term :
an = an−1 + an−2 + an−3
a5 = a4 + a3 + a2 ==> 6 + 3 + 2 ==> 11
6th term :
an = an−1 + an−2 + an−3
a6 = a5 + a4 + a3 ==> 11 + 6 + 3 ==> 20
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