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.

Arithmetic progression

EXERCISE 19.1

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