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

**Question 1. If the n**^{th} term a_{n} of a sequence is given by **a**_{n } = n^{2 }– n +1, write down its first five terms.

^{th}term a

_{n}of a sequence is given by

_{n }= n

^{2 }– n +1, write down its first five terms.

**Solution:**

We have,

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Putting value n = 1 in equation (1), we get

a_{1 }= (1)^{2}– 1 + 1 = 1Putting value n = 2 in equation (1), we get

a_{2}= (2)^{2}– 2 + 1 = 3Putting value n = 3 in equation (1), we get

a_{3}= (3)^{2}– 3 + 1 = 7Putting value n = 4 in equation (1), we get

a_{4}= (4)^{2}– 4 + 1 = 13Putting value n = 5 in equation (1), we get

a_{5}= (5)^{2}– 5 + 1 = 21Hence, the five terms of the given n

^{th}term is1, 3, 7, 13, 21.

**Question 2. A sequence is defined by a**_{n} = n^{3} – 6n^{2} + 11n – 6, n ∈ N. Show that the first three terms of the sequence are zero and all other terms are positive.

_{n}= n

^{3}– 6n

^{2}+ 11n – 6, n ∈ N. Show that the first three terms of the sequence are zero and all other terms are positive.

**Solution:**

Given,

a—(1)_{n }= n^{3 }– 6n^{2 }+ 11n – 6Since n∈ N , therefore first three terms are :

a_{1}= (1)^{3}– 6*(1)^{2}+ 11*(1) – 6 = 0

a_{2}= (2)^{3}– 6*(2)^{2}+ 11*2 – 6 = 0

a_{3}= (3)^{3}– 6*(3)^{2}+11*3 – 6 = 0Hence, the first three terms a1,a2,a3 are zero.

Equation 1 can be rearranged as:

afor n >= 4, an > 0_{n}= (n-2)^{3}– (n-2)Hence, all terms excluding first, second and third are positive.

**Question 3. Find the first four terms of the sequence defined by a**_{1} = 3 and a_{n} = 3a_{n-1} + 2 for all n > 1.

_{1}= 3 and a

_{n}= 3a

_{n-1}+ 2 for all n > 1.

**Solution:**

We have,

aand_{n}= 3a_{n-1}+ 2a_{1}= 3Now,

a_{2 }= 3*a_{1}+ 2 = 3*3 + 2 = 11

a_{3}= 3*a_{2}+ 2 = 3*11 + 2 = 35

a_{4}= 3*a_{3}+ 2 = 3*35 + 2 = 107Hence, the first four terms are

3, 11, 35, 107.

**Question 4. Write the first five terms in each of the following sequences:**

**(i) a**_{1} = 1, a_{n} = a_{n-1} + 2, n > 1

_{1}= 1, a

_{n}= a

_{n-1}+ 2, n > 1

**Solution:**

We have,

aand_{1}= 1a_{n}= a_{n-1}+ 2Now,

a_{2}= a_{1}+ 2 = 1 + 2 = 3

a_{3}= a_{2}+ 2 = 3 + 2 = 5

a_{4}= a_{3}+ 2 = 5 + 2 = 7

a_{5}= a_{4}+ 2 = 7 + 2 = 9Hence, the first five terms are

1, 3, 5, 7, 9.

**(ii) a**_{1} = 1 = a_{2}, a_{n} = a_{n-1} + a_{n-2} , n > 2

_{1}= 1 = a

_{2}, a

_{n}= a

_{n-1}+ a

_{n-2}, n > 2

**Solution:**

We have,

a,_{1}= 1aand_{2}= 1a_{n}= a_{n-1}+ a_{n-2}Therefore,

a_{3}= a_{2}+ a_{1}= 2

a_{4}= a_{3}+ a_{2}= 3

a_{5}= a_{4}+ a_{3}= 5Hence, the first five terms are

1, 1, 2, 3, 5.

**(iii) a**_{1} = a_{2} = 2, a_{n} = a_{n-1} – 1 n > 2

_{1}= a

_{2}= 2, a

_{n}= a

_{n-1}– 1 n > 2

**Solution:**

We have,

aand_{1}= a_{2}= 2a_{n}= a_{n-1}– 1Now,

a_{3}= a_{2}– 1 = 1

a_{4}= a_{3}– 1 = 0

a_{5}= a_{4}– 1 = -1Hence, the first five terms are

2, 2, 1, 0, -1.

**Question 5. The Fibonacci sequence is defined by a**_{1} = 1 = a_{2} , a_{n} = a_{n-1} + a_{n-2} for n > 2. Find a_{n+1}/a_{n} for n = 1, 2, 3, 4, 5.

_{1}= 1 = a

_{2}, a

_{n}= a

_{n-1}+ a

_{n-2}for n > 2. Find a

_{n+1}/a

_{n}for n = 1, 2, 3, 4, 5.

**Solution:**

We have,

a, and_{1}= a_{2}= 1

afor n > 2_{n}= a_{n-1}+ a_{n-2}Now,

a_{3}= a_{2}+ a_{1}= 1 + 1 = 2

a_{4}= a_{3}+ a_{2}= 2 + 1 = 3

a_{5}= a_{4}+ a_{3}= 3 + 2 = 5

a_{6}= a_{5}+ a_{4}= 5 + 3 = 8Therefore,

for n = 1,

a_{n+1}/a_{n}= a_{2}/ a_{1}= 1/1 = 1for n = 2,

a_{n+1}/a_{n}= a_{3}/ a_{2}= 2/1 = 2for n = 3,

a_{n+1}/a_{n}= a_{4}/ a_{3}= 3/2for n = 4,

a_{n+1}/a_{n}= a_{5}/a_{4}= 5/3for n = 5,

a_{n+1}/a_{n}= a_{6}/a_{5}= 8/5Hence,

1, 2, 3/2, 5/3, 8/5are 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,

a,_{1}= 3a,_{2}= -1a,_{3}= -5a_{4}= -9Since,

a_{2}– a_{1}= a_{3}– a_{2}= a_{4}– a_{3}= -4Therefore, It is an A.P with common difference d = -4.

The other three terms are as follows:

a_{5}= -9 + -4 = -13

a_{6}= -13 + -4 = -17

a_{7}= -17 + -4 = -21

**(ii) -1, 1/4, 3/2, 11/4,…**

**Solution:**

We have,

a,_{1}= -1a,_{2}= 1/4a,_{3}= 3/2a_{4}= 11/4Since,

a_{2}– a_{1}= a_{3}– a_{2}= a_{4}– a_{3}= 5/4Therefore, It is an A.P with common difference d = 5/4.

The other three terms are as follows:

a_{5}= 11/4 + 5/4 = 16/4 = 4

a_{6}= 16/4 + 5/4 = 21/4

a_{7}= 21/4 + 5/4 = 26/4 = 13/2

**(iii) √2, 3√2, 5√2, 7√2,…**

**Solution:**

We have,

a,_{1}= √2a,_{2}= 3√2a,_{3}= 5√2a_{4}= 7√2Since,

a_{2}– a_{1}= a_{3}– a_{2}= a_{4}– a_{3}= 2√2Therefore, It is an A.P with common difference d = 2√2.

The other three terms are as follows:

a_{5}= 7√2 + 2√2 = 9√2

a_{6}= 9√2 + 2√2 = 11√2

a_{7}= 11√2 + 2√2 = 13√2

**(iv) 9, 7, 5, 3, …**

**Solution:**

We have,

a,_{1}= 9a,_{2}= 7a,_{3}= 5a_{4}= 3Since,

a_{2}– a_{1}= a_{3}– a_{2}= a_{4}– a_{3}= -2Therefore, It is an A.P with common difference d = -2.

The other three terms are as follows:

a_{5}= 3 + -2 = 1

a_{6}= 1 + -2 = -1

a_{7}= -1 + -2 = -3

**Question 7. The n**^{th} term of a sequence is given by a_{n} = 2n + 7. Show that it is an A.P. Also, find its 7th term.

^{th}term of a sequence is given by a

_{n}= 2n + 7. Show that it is an A.P. Also, find its 7th term.

**Solution:**

We have,

a_{n}= 2n + 7Now,

a_{1}= 2 + 7 = 9

a_{2}= 4 + 7 = 11

a_{3}= 6 + 7 = 13Since,

a_{3}– a_{2}= a_{2}– a_{1}= 2The common difference d = 2, Therefore it is an A.P.

Thus, the 7th term is given by:

a_{7}= 2*7 + 7 = 21.

**Question 8. The n**^{th} term of a sequence is given by a_{n} = 2n^{2} + n + 1. Show that it is not an A.P.

^{th}term of a sequence is given by a

_{n}= 2n

^{2}+ n + 1. Show that it is not an A.P.

**Solution:**

We have,

a_{n}= 2n^{2}+ n + 1Now,

a_{1}= 2*(1)^{2}+ 1 + 1 = 4

a_{2}= 2*(2)^{2}+ 2 + 1 = 11

a_{3}= 2*(3)^{2}+ 3 + 1 = 22Since,

a_{3}– a_{2}≠ a_{2}– a_{1}Therefore, it is not an A.P