Question 1. If the nth term an of a sequence is given by an = n2 – n +1, write down its first five terms.
Solution:
We have, an = n2 – n + 1 —(1)
Putting value n = 1 in equation (1), we get
a1 = (1)2 – 1 + 1 = 1
Putting value n = 2 in equation (1), we get
a2 = (2)2 – 2 + 1 = 3
Putting value n = 3 in equation (1), we get
a3 = (3)2 – 3 + 1 = 7
Putting value n = 4 in equation (1), we get
a4 = (4)2 – 4 + 1 = 13
Putting value n = 5 in equation (1), we get
a5 = (5)2 – 5 + 1 = 21
Hence, the five terms of the given nth term is 1, 3, 7, 13, 21.
Question 2. A sequence is defined by an = n3 – 6n2 + 11n – 6, n ∈ N. Show that the first three terms of the sequence are zero and all other terms are positive.
Solution:
Given, an = n3 – 6n2 + 11n – 6 —(1)
Since n∈ N , therefore first three terms are :
a1 = (1)3 – 6*(1)2 + 11*(1) – 6 = 0
a2 = (2)3 – 6*(2)2 + 11*2 – 6 = 0
a3 = (3)3 – 6*(3)2 +11*3 – 6 = 0
Hence, the first three terms a1,a2,a3 are zero.
Equation 1 can be rearranged as:
an = (n-2)3 – (n-2) for n >= 4, an > 0
Hence, all terms excluding first, second and third are positive.
Question 3. Find the first four terms of the sequence defined by a1 = 3 and an = 3an-1 + 2 for all n > 1.
Solution:
We have, an = 3an-1 + 2 and a1 = 3
Now,
a2 = 3*a1 + 2 = 3*3 + 2 = 11
a3 = 3*a2 + 2 = 3*11 + 2 = 35
a4 = 3*a3 + 2 = 3*35 + 2 = 107
Hence, the first four terms are 3, 11, 35, 107.
Question 4. Write the first five terms in each of the following sequences:
(i) a1 = 1, an = an-1 + 2, n > 1
Solution:
We have, a1 = 1 and an = an-1 + 2
Now,
a2 = a1 + 2 = 1 + 2 = 3
a3 = a2 + 2 = 3 + 2 = 5
a4 = a3 + 2 = 5 + 2 = 7
a5 = a4 + 2 = 7 + 2 = 9
Hence, the first five terms are 1, 3, 5, 7, 9.
(ii) a1 = 1 = a2, an = an-1 + an-2 , n > 2
Solution:
We have, a1 = 1, a2 = 1 and an = an-1 + an-2
Therefore,
a3 = a2 + a1 = 2
a4 = a3 + a2 = 3
a5 = a4 + a3 = 5
Hence, the first five terms are 1, 1, 2, 3, 5.
(iii) a1 = a2 = 2, an = an-1 – 1 n > 2
Solution:
We have,
a1 = a2 = 2 and an = an-1 – 1
Now,
a3 = a2 – 1 = 1
a4 = a3 – 1 = 0
a5 = a4 – 1 = -1
Hence, the first five terms are 2, 2, 1, 0, -1.
Question 5. The Fibonacci sequence is defined by a1 = 1 = a2 , an = an-1 + an-2 for n > 2. Find an+1/an for n = 1, 2, 3, 4, 5.
Solution:
We have,
a1 = a2 = 1, and
an = an-1 + an-2 for n > 2
Now,
a3 = a2 + a1 = 1 + 1 = 2
a4 = a3 + a2 = 2 + 1 = 3
a5 = a4 + a3 = 3 + 2 = 5
a6 = a5 + a4 = 5 + 3 = 8
Therefore,
for n = 1, an+1/an = a2 / a1 = 1/1 = 1
for n = 2, an+1/an = a3 / a2 = 2/1 = 2
for n = 3, an+1/an = a4 / a3 = 3/2
for n = 4, an+1/an = a5/a4 = 5/3
for n = 5, an+1/an = a6/a5 = 8/5
Hence, 1, 2, 3/2, 5/3, 8/5 are the values for n = 1, 2, 3, 4, 5 respectively.
Question 6. Show that each of the following sequences is an A.P. Also, find the common difference and write 3 more terms in each case.
(i) 3, -1, -5, -9, …
Solution:
We have,
a1 = 3, a2 = -1, a3 = -5, a4 = -9
Since,
a2 – a1 = a3 – a2 = a4 – a3 = -4
Therefore, It is an A.P with common difference d = -4.
The other three terms are as follows:
a5 = -9 + -4 = -13
a6 = -13 + -4 = -17
a7 = -17 + -4 = -21
(ii) -1, 1/4, 3/2, 11/4,…
Solution:
We have,
a1 = -1, a2 = 1/4, a3 = 3/2, a4 = 11/4
Since,
a2 – a1 = a3 – a2 = a4 – a3 = 5/4
Therefore, It is an A.P with common difference d = 5/4.
The other three terms are as follows:
a5 = 11/4 + 5/4 = 16/4 = 4
a6 = 16/4 + 5/4 = 21/4
a7 = 21/4 + 5/4 = 26/4 = 13/2
(iii) √2, 3√2, 5√2, 7√2,…
Solution:
We have,
a1 = √2, a2 = 3√2, a3 = 5√2, a4 = 7√2
Since,
a2 – a1 = a3 – a2 = a4 – a3 = 2√2
Therefore, It is an A.P with common difference d = 2√2.
The other three terms are as follows:
a5 = 7√2 + 2√2 = 9√2
a6 = 9√2 + 2√2 = 11√2
a7 = 11√2 + 2√2 = 13√2
(iv) 9, 7, 5, 3, …
Solution:
We have,
a1 = 9, a2 = 7, a3 = 5, a4 = 3
Since,
a2 – a1 = a3 – a2 = a4 – a3 = -2
Therefore, It is an A.P with common difference d = -2.
The other three terms are as follows:
a5 = 3 + -2 = 1
a6 = 1 + -2 = -1
a7 = -1 + -2 = -3
Question 7. The nth term of a sequence is given by an = 2n + 7. Show that it is an A.P. Also, find its 7th term.
Solution:
We have, an = 2n + 7
Now,
a1 = 2 + 7 = 9
a2 = 4 + 7 = 11
a3 = 6 + 7 = 13
Since,
a3 – a2 = a2 – a1 = 2
The common difference d = 2, Therefore it is an A.P.
Thus, the 7th term is given by:
a7 = 2*7 + 7 = 21.
Question 8. The nth term of a sequence is given by an = 2n2 + n + 1. Show that it is not an A.P.
Solution:
We have, an = 2n2 + n + 1
Now,
a1 = 2*(1)2 + 1 + 1 = 4
a2 = 2*(2)2 + 2 + 1 = 11
a3 = 2*(3)2 + 3 + 1 = 22
Since,
a3 – a2 ≠ a2 – a1
Therefore, it is not an A.P