Exercise 8.1 NCERT Class 11 Sequences and Series Best Solutions

Q.1 Q.2 Q.3 Q.4 Q.5 Q.6 Q.7 Q.8 Q.9 Q.10 Q.11 Q.12 Q.13 Q.14

Exercise 8.1 NCERT Class 11 contains 14 questions that are basic and concept-based.

We have written the solutions in such a way that anyone can easily understand the solution steps. Our solutions involve two thought steps. One is knowing the concepts involved, and the other is writing the solution in the required series of steps.

Download NCERT Class 11 Book

Exercise 8.1 NCERT Class 11 Math Solutions

Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:

1. $a_{n}$ $=$ $n (n + 2)$

Thought: To find the first five terms of the given sequence, just put the value of n equal to 1, 2, 3, 4 and 5 in the equation.

Sol:
It is given that n^th term = n(n+2)
Putting n = 1
1^st term = 1×(1+2) = 3
Putting n = 2
2^nd term = 2×(2+2) = 8
Putting n = 3
3^rd term = 3×(3+2) = 15
Putting n = 4
4^th term = 4×(4+2) = 24
Putting n = 5
5^th term = 5×(5+2) = 35
Hence, the first five terms of the given sequence n^th term = n(n+2) are 3, 8, 15, 24 and 35.

2. $a_{n}$ $=$ $\frac{n}{n + 1}$

Sol: It is given that n^th term =

\frac{n}{n+1}

Putting n = 1
1^st term =

\frac{1}{1+1}

\frac{1}{2}

Putting n = 2
2^nd term =

\frac{2}{2+1}

\frac{2}{3}

Putting n = 3
3^rd term =

\frac{3}{3+1}

\frac{3}{4}

Putting n = 4
4^th term =

\frac{4}{4+1}

\frac{4}{5}

Putting n = 5
5^th term =

\frac{5}{5+1}

\frac{5}{6}

Hence, the first five terms of the given sequence n^th term =

\frac{n}{n+1}

are

\frac{1}{2}

\frac{2}{3}

\frac{3}{4}

\frac{4}{5}

\frac{5}{6}

3. $a_{n}$ $=$ $2^{n}$

Sol: It is given that n^th term = 2ⁿ.
Putting n = 1
1^st term = 2¹ = 2
Putting n = 2
2^nd term = 2² = 4
Putting n = 3
3^rd term = 2³ = 8
Putting n = 4
4^th term = 2⁴ = 16
Putting n = 5
5^th term = 2⁵ = 32
Hence, the first five terms of the given sequence n^th term = 2ⁿ are 2, 4, 8, 16 and 32.

4. $a_{n}$ $=$ $\frac{2 n - 3}{6}$

Sol:
It is given that n^th term =

\frac{2n-3}{6}

Putting n = 1
1^st term =

\frac{2×1 – 3}{6}

\frac{-1}{6}

Putting n = 2
2^nd term =

\frac{2×2 – 3}{6}

\frac{1}{6}

Putting n = 3
3^rd term =

\frac{2×3 – 3}{6}

\frac{3}{6}

\frac{1}{2}

Putting n = 4
4^th term =

\frac{2×4 – 3}{6}

\frac{5}{6}

Putting n = 5
5^th term =

\frac{2×5 – 3}{6}

\frac{7}{6}

Hence, the first five terms of the given sequence n^th term = 2ⁿ are

\frac{-1}{6}

\frac{1}{6}

\frac{1}{2}

\frac{5}{6}

and

\frac{7}{6}

5. $a_{n}$ $=$ ${(- 1)}^{n - 1} 5^{n + 1}$

Sol:
It is given that n^th term = (-1)^n-1 5ⁿ⁺¹.
Putting n = 1
1^st term = (-1)^1-1 5¹⁺¹ = (-1)⁰ 5² = 1×25 = 25
Putting n = 2
2^nd term = (-1)^2-1 5²⁺¹ = (-1)¹ 5³ = -1×125 = -125
Putting n = 3
3^rd term = (-1)^3-1 5³⁺¹ = (-1)² 5⁴ = 1×625 = 625
Putting n = 4
4^th term = (-1)^4-1 5⁴⁺¹ = (-1)³ 5⁵ = -1×3125 = -3125
Putting n = 5
5^th term = (-1)^5-1 5⁵⁺¹ = (-1)⁴ 5⁶= 1×15625 = 15625
Hence, the first five terms of the given sequence n^th term = (-1)^n-1 5ⁿ⁺¹ are 25, -125, 625, -3125 and 15625.

6. $a_{n}$ $=$ $n \frac{n^{2} + 5}{4}$

Sol:
It is given that n^th term = $n \frac{n^{2} + 5}{4} .$ Putting n = 1
1^st term = $1 \times \frac{1^{2} + 5}{4}$ = $\frac{6}{4}$ = $\frac{3}{2}$

2^nd term = $2 \times \frac{2^{2} + 5}{4}$ = $2 \times \frac{9}{4}$ = $\frac{9}{2}$

3^rd term = $3 \times \frac{3^{2} + 5}{4}$ = $3 \times \frac{14}{4}$ = $\frac{21}{2}$

4^rd term = $4 \times \frac{4^{2} + 5}{4}$ = $3 \times \frac{21}{4}$ = $\frac{63}{4}$

5^rd term = $5 \times \frac{5^{2} + 5}{4}$ = $5 \times \frac{30}{4}$ = $\frac{75}{2} .$

Hence, the first five terms of the given sequence are $\frac{3}{2},$ $\frac{9}{2},$ $\frac{21}{2},$ $\frac{63}{4},$ and $\frac{75}{2} .$

Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are given:

7. $a_{n}$ $=$ $4 n - 3;$ $a_{17},$ $a_{24}$

Thought: We need to put the value of n = 17 and n = 24 to find the required terms.

Sol: The given sequence is $a_{n} = 4 n - 3$
$a_{17} = 17^{th}$ term = $4 \times 17 - 3 = 68 - 3 = 65$
$a_{24} = 24^{th}$ term = $4 \times 24 - 3 = 96 - 3 = 93$

8. $a_{n} =$ $\frac{n^{2}}{2^{n}};$ $a_{7}$

Sol: The given sequence is

a_{n} =

\frac{n^{2}}{2^{n}}

a_{7} =

7^{th}

term =

\frac{7^{2}}{2^{7}}

\frac{49}{128}

9. $a_{n}$ = ${(-1)}^{n - 1} n^{3}; a_{9}$

Sol: The $n^{th}$ term of the given sequence is ${(-1)}^{n - 1} n^{3}$ .
$a_{9} =$ ${(-1)}^{9 - 1} 9^{3} =$ ${(-1)}^{8} 9^{3} =$ 729

10. $a_{n} = \frac{n (n - 2)}{n + 3}; a_{20 .}$

Sol: The

n^{th}

term of the given sequence is

a_{n} = \frac{n (n - 2)}{n + 3}

a_{20}

\frac{20 (20 - 2)}{20 + 3}

\frac{360}{23} .

Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series:

11. $a_{1} = 3, a_{n} = 3 a_{n - 1} + 2$ , for all n>1.

Thought: First term is given, we can find the next terms using the second expression by putting the first term into it

Sol: Clearly, the first term

(a_{1})

= 3
Now, it is given that

a_{n}

3 a_{n - 1} + 2

,
So, putting the value of n = 2, we get,
The second term =

a_{2}

3 a_{2 - 1} + 2

3 a_{1} + 2

3 \times 3 + 2

= 11.
Similarly,
the third term =

a_{3}

3 a_{3 - 1} + 2

3 a_{2} + 2

3 \times 11 + 2

= 35.
The fourth term =

a_{4}

3 a_{4 - 1} + 2

3 a_{3} + 2

3 \times 35 + 2

= 107.
The fifth term =

a_{5}

3 a_{5 - 1} + 2

3 a_{4} + 2

3 \times 107 + 2

= 323.
Hence, the first five terms of the given sequence are 3, 11, 35, 107 and 323.

12. $a_{1} = - 1, a_{n} =$ $\frac{a_{n - 1}}{n}, n \geq 2$

Sol: Clearly, the first term = -1
Now, it is given that

a_{n} =

\frac{a_{n - 1}}{n}

, which is true for all

n \geq 2

.
So, we can get the value of other terms by just putting the value of n.
The second term =

a_{2} =

\frac{a_{2 - 1}}{2}

\frac{a_{1}}{2}

\frac{- 1}{2}

.
The third term=

a_{3} =

\frac{a_{3 - 1}}{3}

\frac{a_{2}}{3}

\frac{\frac{- 1}{2}}{3}

\frac{- 1}{2 \times 3}

\frac{- 1}{6}

.
The fourth term =

a_{4} =

\frac{a_{4 - 1}}{4}

\frac{a_{3}}{4}

\frac{\frac{- 1}{6}}{4}

\frac{- 1}{6 \times 4}

\frac{- 1}{24}

.
The fifth term =

a_{5} =

\frac{a_{5 - 1}}{5}

\frac{a_{4}}{5}

\frac{\frac{- 1}{24}}{5}

\frac{- 1}{24 \times 5}

\frac{- 1}{120}

.
Hence, the first five terms are

- 1,

\frac{- 1}{2},

\frac{- 1}{6},

\frac{- 1}{24},

and

\frac{- 1}{120} .

13. $a_{1} = a_{2} = 2, a_{n} = a_{n - 1} - 1, n > 2$

Sol: Clearly, the first term = 2,
and the second term = 2,
Now, it is given that $a_{n} = a_{n - 1} - 1,$ so by putting the value of n, we can find the values of other terms.
The third term = $a_{3} = a_{3 - 1} - 1,$ = $a_{2} - 1$ = $2 - 1$ = 1.
The fourth term = $a_{4} = a_{4 - 1} - 1,$ = $a_{3} - 1$ = $1 - 1$ = 0.
The fifth term = $a_{5} = a_{5 - 1} - 1,$ = $a_{4} - 1$ = $0 - 1$ = 0.
Hence the first five terms are 2, 2, 1, 0, and -1.

14. The Fibonacci sequence is defined by
$a_{1} = a_{2} = 1, a n d$
$a_{n} = a_{n - 1} + a_{n - 2}, n > 2$ .
Find $\frac{a_{n + 1}}{a_{n}}$ , for n = 1, 2, 3, 4, 5.

Sol: We need to find the first six terms of the sequence.
Given that $a_{1} = a_{2} = 1$
We need to find the next three terms to find the ratio being asked:
$a_{3} =$ $a_{3 - 1} + a_{3 - 2} =$ $a_{2} + a_{1} = 1 + 1 = 2$
$a_{4} =$ $a_{4 - 1} + a_{4 - 2} =$ $a_{3} + a_{2} = 2 + 1 = 3$
$a_{5} =$ $a_{5 - 1} + a_{5 - 2} =$ $a_{4} + a_{3} = 3 + 2 = 5$
$a_{6} =$ $a_{6 - 1} + a_{6 - 2} =$ $a_{5} + a_{4} = 5 + 3 = 8$
Now,
For n = 1, we have
$\frac{a_{n + 1}}{a_{n}} =$ $\frac{a_{1 + 1}}{a_{1}} =$ $\frac{a_{2}}{a_{1}} =$ $\frac{1}{1} =$ 1.
For n = 2, we get
$\frac{a_{n + 1}}{a_{n}} =$ $\frac{a_{2 + 1}}{a_{2}} =$ $\frac{a_{3}}{a_{2}} =$ $\frac{2}{1} = 2$
For n = 3, we get
$\frac{a_{n + 1}}{a_{n}} =$ $\frac{a_{3 + 1}}{a_{3}} =$ $\frac{a_{4}}{a_{3}} =$ $\frac{3}{2}$
For n = 4, we get
$\frac{a_{n + 1}}{a_{n}} =$ $\frac{a_{4 + 1}}{a_{4}} =$ $\frac{a_{5}}{a_{4}} =$ $\frac{5}{3}$
For n = 5, we get
$\frac{a_{n + 1}}{a_{n}} =$ $\frac{a_{5 + 1}}{a_{5}} =$ $\frac{a_{6}}{a_{5}} =$ $\frac{8}{5}$
Hence, the required values of $\frac{a_{n + 1}}{a_{n}}$ are 1, 2, $\frac{3}{2},$ $\frac{5}{3},$ and $\frac{8}{5}$

Download NCERT Class 11 Book

Exercise 8.1 NCERT Class 11 Math Solutions

Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:

1. a n = n (n+2)

2. a n = n n+1

3. a n = 2 n

4. a n = 2n–3 6

5. a n = (–1) n–1 5 n+1

6. a n = n n 2 + 5 4

Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are given:

7. an = 4n – 3; a17, a24

8. an= n22n; a7

9. an = (-1)n–1n3; a9

10. an=n(n–2)n+3; a20.