Solutions:
(i) an = 3n + 2
Given sequence whose an = 3n + 2
To get the first five terms of given sequence, put n = 1, 2, 3, 4, 5 and we get
a1 = (3 × 1) + 2 = 3 + 2 = 5
a2 = (3 × 2) + 2 = 6 + 2 = 8
a3 = (3 × 3) + 2 = 9 + 2 = 11
a4 = (3 × 4) + 2 = 12 + 2 = 14
a5 = (3 × 5) + 2 = 15 + 2 = 17
∴ the required first five terms of the sequence whose nth term, an = 3n + 2 are 5, 8, 11, 14, 17.
(ii) an = (n – 2)/3
Given sequence whose
On putting n = 1, 2, 3, 4, 5 then can get the first five terms
∴ the required first five terms of the sequence whose nth term,
(iii) an = 3n
Given sequence whose an = 3n
To get the first five terms of given sequence, put n = 1, 2, 3, 4, 5 in the above
a1 = 31 = 3;
a2 = 32 = 9;
a3 = 27;
a4 = 34 = 81;
a5 = 35 = 243.
∴ the required first five terms of the sequence whose nth term, an = 3n are 3, 9, 27, 81, 243.
(iv) an = (3n – 2)/ 5
Given sequence whose
To get the first five terms of the sequence, put n = 1, 2, 3, 4, 5 in the above
And, we get
∴ the required first five terms of the sequence are 1/5, 4/5, 7/5, 10/5, 13/5
(v) an = (-1)n2n
Given sequence whose an = (-1)n2n
To get first five terms of the sequence, put n = 1, 2, 3, 4, 5 in the above.
a1 = (-1)1.21 = (-1).2 = -2
a2 = (-1)2.22 = (-1).4 = 4
a3 = (-1)3.23 = (-1).8 = -8
a4 = (-1)4.24 = (-1).16 = 16
a5 = (-1)5.25 = (-1).32 = -32
∴ the first five terms of the sequence are – 2, 4, – 8, 16, – 32.
(vi) an = n(n – 2)/2
The given sequence is,
To get the first five terms of the sequence, put n = 1, 2, 3, 4, 5.
And, we get
∴ the required first five terms are -1/2, 0, 3/2, 4, 15/2
(vii) an = n2 – n + 1
The given sequence whose, an = n2 – n + 1
To get the first five terms of given sequence, put n = 1, 2, 3, 4, 5.
And, we get
a1 = 12 – 1 + 1 = 1
a2 = 22 – 2 + 1 = 3
a3 = 32 – 3 + 1 = 7
a4 = 42 – 4 + 1 = 13
a5 = 52 – 5 + 1 = 21
∴ the required first five terms of the sequence are 1, 3, 7, 13, 21.
(viii) an = 2n2 – 3n + 1
The given sequence whose an = 2n2 – 3n + 1
To get the first five terms of the sequence, put n = 1, 2, 3, 4, 5.
And, we get
a1 = 2.12 – 3.1 + 1 = 2 – 3 + 1 = 0
a2 = 2.22 – 3.2 + 1 = 8 – 6 + 1 = 3
a3 = 2.32 – 3.3 + 1 = 18 – 9 + 1 = 10
a4 = 2.42 – 3.4 + 1 = 32 – 12 + 1 = 21
a5 = 2.52 – 3.5 + 1 = 50 – 15 + 1 = 36
∴ the required first five terms of the sequence are 0, 3, 10, 21, 36.
(ix) an = (2n – 3)/ 6
Given sequence whose,
To get the first five terms of the sequence we put n = 1, 2, 3, 4, 5.
And, we get
∴ the required first five terms of the sequence are -1/6, 1/6, 1/2, 5/6 and 7/6