Arithmetic progression

EXERCISE 19.1

QUESTION 1

If the n^{\text { th }} term a_{n} of a sequence is given by a_{n}=n^{2}-n+1, write down its first five terms.

Sol :

Given : a_{n}=n^{2}-n+1

For n=1, a_{1}=1^{2}-1+1

= 1

For n=2, a_{2}=2^{2}-2+1

= 2

For n=3, a_{3}=3^{2}-3+1

= 3

For n=4, a_{4}=4^{2}-4+1

= 13

For n=5, a_{5}=5^{2}-5+1

= 21

Thus, the first five terms of the sequence are 1,3,7,13,21

 

QUESTION 2

A sequence is defined by a_{n}=n^{3}-6 n^{2}+11 n-6, n \in N Show that the first three terms of
the sequence are zero and all other terms are positive.

Sol :

Given :

a_{n}=n^{3}-6 n^{2}+11 n-6, n \in N 

For n=1, a_{1}=1^{3}-6 \times 1^{2}+11 \times 1-6=0

For n=2, a_{2}=2^{3}-6 \times 2^{2}+11 \times 2-6=0

For n=3, a_{3}=3^{3}-6 \times 3^{2}+11 \times 3-6=0

For n=4, a_{4}=4^{3}-6 \times 4^{2}+11 \times 4-6=6>0

For n=5, a_{5}=5^{3}-6 \times 5^{2}+11 \times 5-6=24>0

and so on

Thus, the first three terms are zero and the rest of the terms are positive in the sequence .

 

QUESTION 3

Let <a_{n}> be a sequence defined by a_{1}=3 and, a_{n}=3 a_{n-1}+2, for all n>1 .Find the first four terms of the sequence.

Sol :

Given :

a_{1}=3

And, a_{n} =3 a_{n-1}+2 for all n>1

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

a_{3} =3 a_{3-1}+2 =3 a_{2}+2 =35

a_{4} =3 a_{4-1}+2 =3 a_{3}+2 =107

Thus, the first four terms of the sequence are 3,11,35,107

 

QUESTION 4

Let <a_{n}> be a sequence. Write the first five terms in each of the following:

(i) a_{1}=1, a_{n}=a_{n-1}+2, n \geq 2

Sol :

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

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

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

a_{5}=a_{4}+2=9

Hence, the five terms are 1,3,5,7 and 9 .

 

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

Sol :

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

Hence, the five terms are 1,1,2,3 and 5

 

(iii) a_{1}=a_{2}=2, a_{n}=a_{n-1}-1, n>2

Sol :

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

a_{4}=a_{3}-1=1-1=0

a_{5}=a_{4}-1=0-1=-1

Hence, the five terms are 2,2,1,0 and -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 \dfrac{a_{n+1}}{a_{n}} for n=1,2,3,4,5

Sol :

a_{1}=1=a_{2}, a_{n}=a_{n-1}+a_{n-2} for n>2

Then, we have:

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=8

For n=1, \dfrac{a_{n+1}}{a_{n}}=\dfrac{a_{2}}{a_{1}}=\dfrac{1}{1}=1

For n=2, \dfrac{a_{n+1}}{a_{n}}=\dfrac{a_{3}}{a_{2}}=\dfrac{2}{1}=2

For n=3, \dfrac{a_{n+1}}{a_{n}}=\dfrac{a_{4}}{a_{3}}=\dfrac{3}{2}

For n=4, \dfrac{a_{n+1}}{a_{n}}=\dfrac{a_{5}}{a_{4}}=\dfrac{5}{3}

For n=5, \dfrac{a_{n+1}}{a_{n}}=\dfrac{a_{6}}{a_{5}}=\dfrac{8}{5}

 

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 \ldots

Sol :

-1-3=-4

-5-(-1)=-4

-9-(-5)=-4 \ldots

Thus, the sequence is an A.P. with the common difference being -4 .
The next three terms are as follows:

-9-4=-13

-13-4=-17

-17-4=-21

 

(ii) -1,1 / 4,3 / 2,11 / 4, \dots

Sol :

we have

1 / 4-(-1)=5 / 4

3 / 2-1 / 4=5 / 4

11 / 4-3 / 2=5 / 4

Thus, the sequence is an A.P. with the common difference being (5 / 4) The next three terms are as follows:

11 / 4+5 / 4=16 / 4=4

16 / 4+5 / 4=21 / 4

21 / 4+5 / 4=26 / 4

 

(iii) \sqrt{2}, 3 \sqrt{2}, 5 \sqrt{2}, 7 \sqrt{2}, \ldots

Sol :

3 \sqrt{2}-\sqrt{2}=2 \sqrt{2}

5 \sqrt{2}-3 \sqrt{2}=2 \sqrt{2}

7 \sqrt{2}-5 \sqrt{2}=2 \sqrt{2}

Thus, the sequence is an A.P. with the common difference being (2 \sqrt{2}) The next three terms are as follows:

7 \sqrt{2}+2 \sqrt{2}=9 \sqrt{2}

9 \sqrt{2}+2 \sqrt{2}=11 \sqrt{2}

11 \sqrt{2}+2 \sqrt{2}=13 \sqrt{2}

 

(iv) 9,7,5,3, \dots

Sol :

7-9=-2

5-7=-2

3-5=-2

Thus, the sequence is an A.P. with the common difference being (-2) The next three terms are as follows:

3-2=1

1-2=-1

-1-2=-3

 

QUESTION 7

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

Sol :

a_{n}=2 n+7

a_{1}=2 \times 1+7=9

a_{2}=2 \times 2+7=11

a_{3}=2 \times 3+7=13

a_{4}=2 \times 4+7=15

and so on

So, common difference (d)=11-9=2

Thus, the above sequence is an A . P . with the common difference as 2
a_{7}=2 \times 7+7=21

a_{7}=2 \times 7+7=21

 

QUESTION 8

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

Sol :

We have :

a_{1}=2 \times 1^{2}+1+1

= 4

a_{2}=2 \times 2^{2}+2+1

= 11

a_{3}=2 \times 3^{2}+3+1

= 22

a_{2}-a_{1}=11-4

= 7

and a_{3}-a_{2}=22-11

= 11

Since, a_{2}-a_{1} \neq a_{3}-a_{2}

Hence, it is not an AP.

 

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