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• RD Sharma Class 11 Solutions for Maths

# Class 11 RD Sharma Solutions – Chapter 19 Arithmetic Progressions- Exercise 19.1

### Question 1. If the nth term an of a sequence is given byan  = 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.

### (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.

### (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