Exercise 5.1

1.In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?

(i) The taxi fare after each km when the fare is ` 15 for the first km and ` 8 for each additional km.

Sol.taxi fare for 1 km

=Rs $$15=a_1$$

A.T.Q.;Taxi fare for 2 km

=$$Rs 15+Rs 8 =Rs 23=a_2$$

Taxi fare for 3 km

=$$Rs 23+Rs8=Rs 31=a_3$$

Taxi fare for 4 km

=$$Rs 31+Rs 8=Rs 39=a_4$$

And so on .

$$a_2-a_1=Rs 23-Rs 15=Rs 8$$

$$a_3-a_2=Rs 31-Rs 23=Rs 8$$

$$a_4-a_3=Rs 39-Rs 31=Rs 8$$

i.e., $$a_k+1-a_k$$ is the same everytime.

So, this list of numbers forms an arithmetic progression with the first term a=rs 15 and the common different d=rs 8.

(ii) The amount of air present in a cylinder when a vacuum pump removes  1/ 4  of the air remaining in the cylinder at a time.

Sol.Amount of air present in the cylinder =x units (say)=$$a_1$$

ATQ.;amount of air present in the cylinder after one time removal of air by the vacuum pump =$$x-\frac{x}{4}=\frac{3x}{4} \:\:units =a_2$$

Amount of air present in the cylinder after two times removal of air by the vaccum

Pump =$$\frac{3x}{4}-\frac{1}{4}(\frac{3x}{4})=\frac{3x}{4}-\frac{3x}{16}=\frac{9x}{16}$$ units =$$a_3$$.

Amount of air present in the cylinder after three times removal of air by the vacuum pump

=$$\frac{9x}{16}-\frac{1}{4}(\frac{9x}{16})=\frac{9x}{16}-\frac{9x}{64}$$

=$$\frac{36x-9x}{64}=\frac{27x}{64} \:\: units =a_4$$

and so on.

$$a_2-a_1=\frac{3x}{4}-x$$

=$$\frac{3x-4x}{4}=-\frac{x}{4}$$

$$a_3-a_2=\frac{9}{16}x-\frac{3}{4}x$$

=$$\frac{9x-12x}{16}=\frac{-3x}{16}units$$

As $$a_2-a_1≠a_3-a_2$$

This list of numbers does not form an AP.

(iii) The cost of digging a well after every metre of digging, when it costs ` 150 for the first metre and rises by ` 50 for each subsequent metre.

Sol.cost of digging the well after 1 meter of digging =Rs 150=$$a_1$$

Cost of digging the well after 2 meters of digging

=$$Rs 150+Rs 50=Rs200=a_2$$

Cost of digging the well after 3 metres of digging

=$$Rs 200+ Rs 50 =Rs 250=a_3$$

Cost of digging the well after 4 metres of digging

=$$Rs 250+Rs 50=Rs 300=a_4$$

And so on.

$$a_2-a_1=Rs 200-Rs150=Rs50$$

$$a_3-a_2=Rs 250-Rs 200=Rs 50$$

$$a_4-a_3=Rs 300-Rs 250=Rs 50$$

i.e., $$a_{k+1}$$ is the same everytime. So this list of numbers forms an AP with the first term a=Rs 150 and the common different d=Rs 50.

(iv) The amount of money in the account every year, when ` 10000 is deposited at compound interest at 8 % per annum.

Sol.we know that amount of present value P at r% compound interest after n years

=$$P(1+\frac{r}{100})^{n}$$

Amount of money after 1 year

=$$Rs 10000 (1+\frac{8}{100})=a_1$$

Amount of money after 2 years

=$$Rs 10000(1+\frac{8}{100})^{2}=a_2$$

Amount of money after 3 years

=$$Rs 10000(1+\frac{8}{100})^{3}=a_3$$

Amount of money after 4 years

=$$Rs 10000(1+\frac{8}{100})^{4}=a_4$$

And so on.

$$a_2-a_1$$

=$$Rs 10000(1+\frac{8}{100})^{2}-Rs 10000(1+\frac{8}{100})$$

=$$Rs 10000(1+\frac{8}{100})(1+\frac{8}{100}-1)$$

=$$Rs 10000 (1+\frac{8}{100})(\frac{8}{100})$$

$$a_3-a_2$$

=$$Rs 10000(1+\frac{8}{100})^{2}-Rs 10000(1+\frac{8}{100})^{2}$$

=$$Rs 10000(1+\frac{8}{100})^{2}(1+\frac{8}{100}-1)$$

=$$Rs 10000(1+\frac{8}{100})^{2}(\frac{8}{100})$$

As

$$a_2-a_1≠a_3-a_2,$$

This list of numbers does not form an AP.

2.Write first four terms of the AP, when the first term a and the common difference d are given as follows:

(i) a = 10, d = 10

Sol.first term   =a=10

Second term =10+d=10+10=20

Third term = 20+d=20+10=30

Fourth term = 30+d=30+10=40

Hence, first four terms of the given AP

Are

10, 20, 30, 40

(ii) a = –2, d = 0

Sol.first term   =a=-2

Second term =-2+d=-2+0=-2

Third term =-2+d=-2+0=-2

Fourth term =-2+d=2+0=-2

Hence, first four terms of the given AP are -2, -2, -2 ,-2 respectively.

(iii) a = 4, d = – 3

Sol.First term =a=4

Second term =4+d=4+(-3)=1

Third term =1+d=1+(-3)=-2

Fourth term  =-2+d= -2+(-3)

=-5

Hence, four first terms of the given AP are 4, 1, -2, -5.

(iv) a = – 1, d =  1/ 2

Sol.first term = a=-1

Second term =$$-1+d=-1+\frac{1}{2}=-\frac{1}{2}$$

Third term =$$-\frac{1}{2}+d=-\frac{1}{2}+\frac{1}{2}=0$$

Fourth term =$$0+d=0+\frac{1}{2}=\frac{1}{2}$$

(v) a = – 1.25, d = – 0.25

Sol.first term  =a=-1.25

Second term  =-1.25+d

=-1.25+(-0.25)

=-1.50

Third term  =-1.50+d

=-1.50+(-.025)

=-1.75

Fourth term =-1.75+d

=-1.75+(-0.25)

=-2.00

Hence first four terms of the given AP are -1.25, -1.75, -2.00.

3.For the following APs, write the first term and the common difference:

(i) 3, 1, – 1, – 3, . . .

Sol.first term (a)=3

Common difference (d)

=$$a_2-a_1=1-3=-2$$

(ii) – 5, – 1, 3, 7, . . .

Sol.First term (a)=-5

Common difference(d)

=$$a_2-a_1=-1-(-5)$$

=-1+5=4

iii)$$\frac{1}{3},\frac{5}{3},\frac{9}{3}, \frac{13}{3},$$….

First term $$(a)=\frac{1}{3}$$

common difference (d)

=$$a_2-a_1=\frac{5}{3}-\frac{1}{3}=\frac{4}{3}$$

iv)0.6, 1.7, 2.8, 3.9,…

Sol.first term (a)=0.6

common difference $$(d)=a_2-a_1$$

=1.7-0.6=1.1

4.Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.

(i) 2, 4, 8, 16, . .

Sol.$$a_2-a_1=4-2=2$$

$$a_3-a_2=8-4=4$$

$$a_4-a_3=16-8=8$$

As, $$a_2-a_1≠a_3-a_2$$

So, the given list of numbers does not form an AP.

ii)2, $$\frac{5}{2},3, \frac{7}{2},$$…..

$$a_2-a_1=\frac{5}{2}-2=\frac{1}{2}$$

$$a_3-a_2=3-\frac{5}{2}=\frac{1}{2}$$

$$a_4-a_3=\frac{7}{2}-3=\frac{1}{2}$$

i.e., $$a_{k+1}-a_k$$ is the same every time. So, the given list of numbers forms an A.P with the common difference $$d=\frac{1}{2}$$

the next three terms are:

$$\frac{7}{2}+\frac{1}{2}=4, 4+\frac{1}{2}=\frac{9}{2}$$

And $$\frac{9}{2}+\frac{1}{2}=5$$

iii)-1.2, -3.2, -5.2, -7.2……

$$a_2-a_1=-3.2-(-1.2)$$

=-3.2+1.2

=-2.0

$$a_3-a_2=-5.2-(-3.2)$$

=-5.2+3.2

=-2.0

$$a_4-a_3=-7.2-(-5.2)$$

=-7.2+5.2

=-2.0

i.e., $$a_{k+1}$$ is the same everytime.

So, the given list of numbers forms an AP with common different d=-2.0

The next three terms are:

-7.2+(-2.0)=-9.2

-9.2+(-2.0)=-11.2

-11.2+(-2.0)=-13.2

iv)-10, -6, -2, 2,….

$$a_2-a_1=-6-(-10)$$

=-6+10=4

$$a_3-a_2=-2-(-6)$$

=-2+6=4

$$a_4-a_3=2-(-2)$$

=2+2=4

So, the given list of numbers forms an AP with common different d=4

The next three terms are:

2+4=6, 6+4=10

10+4=14.

v)3, $$3+\sqrt{2}, 3+2\sqrt{2}, 3+3\sqrt{2}$$…..

$$a_2-a_1=(3+\sqrt{3})-3=\sqrt{2}$$

$$a_3-a_2=(3+2\sqrt{2})-(3+\sqrt{2})$$

=$$2\sqrt{2}-\sqrt{2}=(2-1)\sqrt{2}=\sqrt{2}$$

$$a_4-a_3=(3+3\sqrt{2})-(3+2\sqrt{2})$$

=$$3\sqrt{2}-2\sqrt{2}=(3-2)\sqrt{2}=\sqrt{2}$$

Ie., $$a_{k+1} -a_k$$  is the same every time.

So, the given list of numbers forms an AP with the common difference $$d=\sqrt{2}$$

The next three terms are :

$$(3+3\sqrt{2})+\sqrt{2}=3+4\sqrt{2}$$,

$$(3+4\sqrt{2})+\sqrt{2}=3+5\sqrt{2}$$,

$$(3+5\sqrt{2})+\sqrt{2}=3+6\sqrt{2}$$,

vi)0.2, 0.22, 0.222, 0.2222,…

$$a_2-a_1=0.22-0.2=0.02$$

$$a_3-a_2=0.222-0.22=0.002$$

$$a_2-a_1≠a_3-a_2$$, the given list of numbers does not form an AP.

vii)0,  -4,  -8, -12,…

$$a_2-a_1=-4-0 =-4$$

$$a_3-a_2=-8-(-4)  = -8+4=-4$$

$$a_4-a_3=-12-(-8)=-12+8=-4$$

i.e.:$$a_{k+1} -a_k$$ is the same everytime.

So, the given list of numbers forms an AP with the common difference d=-4.

The next three terms are:

-12+(-4)=-12-4=-16,

-16+(-4)=-16-4=-20

-20+(-4)=-20-4=-24

viii)$$-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}$$,

$$a_2-a_1=-\frac{1}{2}-(-\frac{1}{2})=-\frac{1}{2}+\frac{1}{2}=0$$

$$a_3-a_2=-\frac{1}{2}-(-\frac{1}{2})=-\frac{1}{2}+\frac{1}{2}=0$$

$$a_4-a_3=-\frac{1}{2}-(-\frac{1}{2})=-\frac{1}{2}+\frac{1}{2}=0$$

i.e.,$$a_{k+1} -a_k$$ is the same everytime.

So, the given list of numbers forms an AP with the common difference d=0.

The next three terms are:

-$$\frac{1}{2}+0=-\frac{1}{2}$$,  $$-\frac{1}{2}+0=-\frac{1}{2}$$

And

$$-\frac{1}{2}+0=-\frac{1}{2}$$

ix)1, 3, 9, 27,…..

$$a_2-a_1=3-1=2$$

$$a_3-a_2=9-3=6$$

$$a_2-a_1≠a_3-a_2$$, the given list of numbers does not form an AP.

x)a, 2a, 3a, 4a…..

$$a_2-a_1=2a-a  =a$$

$$a_3-a_2=3a-2a=a$$

$$a_4-a_3=4a-3a=a$$

i.e., $$a_{k+1} -a_k$$ is the same everytime.so, the given list of numbers forms an AP with the common difference d=a.

the next three terms are:

4a+a=5a, 5a+a=6a

6a+a=7a.

xi)a, $$a^{2}, a^{3}, a^{4}$$,…

$$a_2-a_1=a^{2}-a=a(a-1)$$

$$a_3 -a_2=a^{3}-a^{2}=a^{2}(a-1)$$

As $$a_2-a_1≠a_3-a_2$$, the given list of numbers does not form an AP.

xii)$$\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}$$,…

$$a_2-a_1=\sqrt{8}-\sqrt{2}=\sqrt{4X2}-\sqrt{2}$$

=$$2\sqrt{2}-\sqrt{2}=\sqrt{2}(2-1)=\sqrt{2}$$

$$a_3-a_2=\sqrt{18}-\sqrt{8}=\sqrt{9X2}-\sqrt{4X2}$$

=$$3\sqrt{2}-2\sqrt{2}=\sqrt{2}(3-2)=\sqrt{2}$$

$$a_4-a_3=\sqrt{32}-\sqrt{18}$$

=$$\sqrt{16X2}-\sqrt{9X2}$$

=$$\sqrt{16}\sqrt{2}-\sqrt{9}\sqrt{2}$$

=$$4\sqrt{2}-3\sqrt{2}$$

=$$(4-3)\sqrt{2}=\sqrt{2}$$

i.e., $$a_{k+1} ,-a_k$$ is the same everytime. So, the given list of numbers form an AP with the common difference $$d=\sqrt{2}$$

the next three terms are:

$$\sqrt{32}+\sqrt{2}=4\sqrt{2}+\sqrt{2}=5\sqrt{2}$$

=$$\sqrt{25}\sqrt{2}=\sqrt{25X2}=\sqrt{50},$$

$$5\sqrt{2}+\sqrt{2}=6\sqrt{2}=\sqrt{36} X \sqrt{2}$$

=$$\sqrt{36 X 2}=\sqrt{72}$$

And $$6\sqrt{2}+\sqrt{2}=7\sqrt{2}=\sqrt{49}\sqrt{2}$$

$$\sqrt{49 X 2}=\sqrt{98}$$

xiii) $$\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}$$,….

$$a_2-a_1=\sqrt{6}-\sqrt{3}=\sqrt{3X2}-\sqrt{3}$$

=$$\sqrt{3}\sqrt{2}-\sqrt{3}=\sqrt{3}(\sqrt{2}-1)$$

$$a_3-a_2=\sqrt{9}-\sqrt{6}=\sqrt{3X3}-\sqrt{3X2}$$

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

=$$\sqrt{3}(\sqrt{3}-\sqrt{2})$$

As $$a_2-a_1≠a_3-a_2$$, the given list of numbers does not form an AP.

(xiv)$$1^{2}, 3^{2}, 5^{2}, 7^{2}$$,….

$$a_2-a_1=3^{2}-1^{2}=9-1=8$$

$$a_3-a_2=5^{2}-3^{2}=25-9=16$$

as $$a_2-a_1≠a_3-a_2$$, the given list of numbers does not form an AP>

xv)$$1^{2}, 5^{2}, 7^{2}, 73,$$…

$$a_2-a_a=5^{2}-1^{1}$$

=25-1=24

$$a_3-a_2=7^{2}-5^{2}$$

=49-25=24

$$a_4-a_3=73-7^{2}=73-49=24$$

i.e.,$$ a_{k+1}$$, $$ -a_k$$ is the same everytime.

So, the given list of numbers forms an AP with the common difference d=24.

The next three terms are:

73+24=97, 97+24=121

And  121+24=145.

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