NCERT Solutions for Class 11 Maths Chapter 8 Sequences and Series Ex 8.1
Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:
Ex 8.1 Class 11 Maths Question 1.
an = n(n + 2)
Solution:
Given, an = n(n + 2)
Substituting n = 1, 2, 3, 4 and 5, we get
a1 = 1(1 + 2) = 1 × 3 = 3
a2 = 2(2 + 2) = 2 × 4 = 8
a3 = 3(3 + 2) = 3 × 5 = 15
a4 = 4(4 + 2) = 4 × 6 = 24
a5 = 5(5 + 2) = 5 × 7 = 35
∴ The first five terms of the give sequence are 3, 8, 15, 24 and 35, respectively.
Ex 8.1 Class 11 Maths Question 2.
an = n/(n+1)
Solution:
Ex 8.1 Class 11 Maths Question 3.
an = 2n
Solution:
Ex 8.1 Class 11 Maths Question 4.
an = (2n−3)/6
Solution:
Ex 8.1 Class 11 Maths Question 5.
an = (-1)n-1 5n+1
Solution:
Given, an = (-1)n-1 5n+1
Substituting n = 1, 2, 3, 4 and 5, we get
a1 = (-1)1-1 51+1 = (-1)° 52 = 25
a2 = (-1)2-1 52+1 = (-1)1 53 = -125
a3 = (-1)3-1 53+1 = (-1)2 54 = 625
a4 = (-1)4-1 54+1 = (-1)3 55 = -3125
a5 = (-1)5-1 55+1 = (-1)4 56 = 15625
∴ The first five terms of the given sequence are 25, -125, 625, -3125 and 15625, respectively.
Ex 8.1 Class 11 Maths Question 6.
an = n(n2 + 5)/4
Solution:
Given, an = n(n2 + 5)/4
Substituting n = 1, 2, 3, 4 and 5, we get
Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:
Ex 8.1 Class 11 Maths Question 7.
an = 4n – 3; a17, a24
Solution:
Given, an = 4n – 3
Substituting n = 17, we get a17 = 4(17) – 3 = 68 – 3 = 65
Substituting n = 24, we get a17 = 4(24) – 3 = 96 – 3 = 93
Ex 8.1 Class 11 Maths Question 8.
an = n2/2n; a7
Solution:
Given, an = n2/2n
Substituting n = 7, we get a7 = (7)2/27 = 49/128
Ex 8.1 Class 11 Maths Question 9.
an = (-1)n – 1 n3; a9
Solution:
Given, an = (-1)n – 1 n3
Substituting n = 9, we get a9 = (-1)9 – 1 93 = (-1)8 × 729 = 729
Ex 8.1 Class 11 Maths Question 10.
an = n(n−2)/(n+3); a20
Solution:
Given, an = n(n−2)/(n+3)
Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series:
Ex 8.1 Class 11 Maths Question 11.
a1 = 3, an = 3an-1 + 2 for all n > 1
Solution:
Given, a1 = 3, an = 3an-1 + 2
⇒ a1 = 3, a2 = 3a1 + 2 = 3 × 3 + 2 = 9 + 2 = 11,
a3 = 3a2 + 2 = 3 × 11 + 2 = 33 + 2 = 35,
a4 = 3a3 + 2 = 3 × 35 + 2 = 105 + 2 = 107,
a5 = 3a4 + 2 = 3 × 107 + 2 = 321 + 2 = 323
Hence, the first five terms of the given sequence are 3, 11, 35, 107 and 323, respectively.
The corresponding series is 3 + 11 + 35 + 107 + 323 + ………..
Ex 8.1 Class 11 Maths Question 12.
a1 = -1, an = an−1/n, n ≥ 2
Solution:
Given,
Hence, the first five terms of the given sequence are –1, -1/2, -1/6, -1/24 and -1/120, respectively.
The corresponding series is:
Ex 8.1 Class 11 Maths Question 13.
Solution:
Given, a1 = a2 = 2, an = an-1 – 1, n > 2
a1 = 2, a2 = 2, a3 = a2 – 1 = 2 – 1 = 1,
a4 = a3 – 1 = 1 – 1 = 0 and a5 = a4 – 1 = 0 – 1 = -1
Hence, the first five terms of the given sequence are 2, 2, 1, 0 and -1, respectively.
The corresponding series is 2 + 2 + 1 + 0 + (-1) + ……
Ex 8.1 Class 11 Maths Question 14.
The Fibonacci sequence is defined by 1 = a1 = a2 and an = an-1 + an-2, n > 2.
Find an+1/an, for n = 1, 2, 3, 4, 5
Solution:
We have,
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