Chapter 3:Pair of Linear Equations in Two Variables

Chapter 5: Arithmetic Progression

Chapter 8: Introduction to Trigonometry

Chapter 9:Some Applications of Trigonometry

Chapter 12:Area Related to Circles

Arithmetic Progression

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

According to question

the fare for journey of first 1 km = Rs 15

Taxi fare for next 2 kms = 15+8 = 23

Taxi fare for next 3 kms = 23+8 = 31

Taxi fare for next 4 kms = 31+8 = 39

15, 23, 31, 39 ......

Here common difference is 8 , So, it is an A P

(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

Let the volume of air in a cylinder, is *V* litres.

The vacuum pump removes 1/4th of air remaining in the cylinder at a time.

So

1-1/4 = 3/4th
part of air will remain.

So, volumes will be *V*, 3*V*/4 , (3*V*/4)^{2} ,
(3*V*/4)^{3}........

d= 3y/4 - y = - y/4

d= 9y/4 - 3y/4 = -3y/4

Since common difference between them is not same . Therefore, this series is not an A.P.

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

According to question

The Cost of digging a well for first metre = Rs.150

Cost of digging a well for Second metres = Rs.150+50 = Rs.200

Cost of digging a well for third metres = Rs.200+50 = Rs.250

Cost of digging a well for fourth metres =Rs.250+50 = Rs.300

And so on..

hence 150, 200, 250, 300 … forms an A.P. with a common difference of 50 between each term.

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

We know that if Rs. P is deposited at *r*% compound interest per annum for n years, the amount
of money will be:

P(1+r/100)^{n}

So, the amount of money in the account in the first years ,second year third year and so on....;

10000(1+8/100), 10000(1+8/100)^{2}, 10000(1+8/100)^{3}……

So the terms of this series do not have the common difference between them. Therefore, this is not an A.P.

(i) *a* = 10, *d* = 10

(ii) *a* = -2, *d* = 0

(iii) *a* = 4, *d* = – 3

(iv) *a* = -1 *d* = 1/2

(v) *a* = – 1.25, *d* = – 0.25

(i) *a* = 10, *d* = 10

Let the Arithmetic Progression series be *a _{1}, a_{2}, a_{3},
a_{4}, a_{5} …*

*a*_{1} = *a* = 10

*a*_{2} = *a*_{1}+*d* = 10+10 = 20

*a*_{3} = *a*_{2}+*d* = 20+10 = 30

*a*_{4} = *a*_{3}+*d* = 30+10 = 40

*a*_{5} = *a*_{4}+*d* = 40+10 = 50

And so on…

So the A.P. series will be 10, 20, 30, 40, 50 …

And First four terms of an A.P. will be 10, 20, 30, and 40.

(ii) *a* = – 2, *d* = 0

Let the Arithmetic Progression series be *a _{1}, a_{2}, a_{3},
a_{4}, a_{5} …*

*a*_{1} = *a* = -2

*a*_{2} = *a*_{1}+*d* = – 2+0 = – 2

*a*_{3} = *a*_{2}+d = – 2+0 = – 2

*a*_{4} = *a*_{3}+*d* = – 2+0 = – 2

Sothe A.P. series will be – 2, – 2, – 2, – 2 …

And, First four terms of an A.P. will be – 2, – 2, – 2 and – 2.

(iii) *a* = 4, *d* = – 3

Let the Arithmetic Progression series be *a _{1}, a_{2}, a_{3},
a_{4}, a_{5} …*

*a*_{1} = *a* = 4

*a*_{2} = *a*_{1}+*d* = 4-3 = 1

*a*_{3} = *a*_{2}+*d* = 1-3 = – 2

*a*_{4} = *a*_{3}+*d* = -2-3 = – 5

So the A.P. series will be 4, 1, – 2 – 5 …

And, first four terms of an A.P. will be 4, 1, – 2 and – 5.

(iv) *a* = – 1, *d* = 1/2

Let the Arithmetic Progression series be *a _{1}, a_{2}, a_{3},
a_{4}, a_{5} …*

*a*_{2} = *a*_{1}+*d* = -1+1/2 = -1/2

*a*_{3} = *a*_{2}+*d* = -1/2+1/2 = 0

*a*_{4} = *a*_{3}+*d* = 0+1/2 = 1/2

So the A.P. series will be-1, -1/2, 0, 1/2

And First four terms of an A.P. will be -1, -1/2, 0 and 1/2.

(v) *a* = – 1.25, *d* = – 0.25

Let , the Arithmetic Progression series be *a _{1}, a_{2}, a_{3},
a_{4}, a_{5} …*

*a*_{1} = *a* = – 1.25

*a*_{2} = *a*_{1} + *d* = – 1.25-0.25 = – 1.50

*a*_{3} = *a*_{2} + *d* = – 1.50-0.25 = – 1.75

*a*_{4} = *a*_{3} + *d* = – 1.75-0.25 = – 2.00

So, the A.P series will be 1.25, – 1.50, – 1.75, – 2.00 ……..

And first four terms of an A.P. will be – 1.25, – 1.50, – 1.75 and – 2.00.

(i) 3, 1, – 1, – 3 …

(ii) -5, – 1, 3, 7 …

(iii) 1/3, 5/3, 9/3, 13/3 ….

(iv) 0.6, 1.7, 2.8, 3.9 …

3, 1, – 1, – 3 …

First term, *a* = 3

Common difference, *d* = 2nd term – 1st term

⇒ 1 – 3 = -2

⇒ d = -2

(ii)given – 5, – 1, 3, 7 …

First term, *a* = -5

Common difference, *d* = 2nd term – 1st term

⇒ ( – 1)-( – 5) = – 1+5 = 4

(iii) Given 1/3, 5/3, 9/3, 13/3 ….

First term, *a* = 1/3

Common difference, *d* = 2nd term – 1st term

⇒ 5/3 – 1/3 = 4/3

(iv) Given 0.6, 1.7, 2.8, 3.9 …

First term, *a* = 0.6

Common difference, *d* = 2nd term – 1st term

⇒ 1.7 – 0.6

⇒ 1.1

(i) 2, 4, 8, 16 …

(ii) 2, 5/2, 3, 7/2 ….

(iii) -1.2, -3.2, -5.2, -7.2 …

(iv) -10, – 6, – 2, 2 …

(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2

(vi) 0.2, 0.22, 0.222, 0.2222 ….

(vii) 0, – 4, – 8, – 12 …

(viii) -1/2, -1/2, -1/2, -1/2 ….

(ix) 1, 3, 9, 27 …

(x) *a*, 2*a*, 3*a*, 4*a* …

(xi) *a*, *a*^{2}, *a*^{3}, *a*^{4} …

(xii) √2, √8, √18, √32 …

(xiii) √3, √6, √9, √12 …

(xiv) 1^{2}, 3^{2}, 5^{2}, 7^{2} …

(xv) 1^{2}, 5^{2}, 7^{2}, 7^{3} …

(i) here ,

2, 4, 8, 16 …

Here, the common difference is;

*a*_{2} – *a*_{1} = 4 – 2 = 2

*a*_{3} – *a*_{2} = 8 – 4 = 4

*a _{4}* –

Since, common difference is not the same every time.

hence the given series are not forming an A.P.

(ii) Given, 2, 5/2, 3, 7/2 ….

Here,

*a*_{2} – *a*_{1} = 5/2-2 = 1/2

*a*_{3} – *a*_{2} = 3-5/2 = 1/2

*a*_{4} – *a*_{3} = 7/2-3 = 1/2

here the common difference is same every time.

Since *d* = 1/2 and the given series are in A.P.

The next three terms

*a*_{5} = 7/2+1/2 = 4

*a*_{6} = 4 +1/2 = 9/2

*a*_{7} = 9/2 +1/2 = 5

(iii) Given, -1.2, – 3.2, -5.2, -7.2 …

Here,

*a*_{2} – *a*_{1} = (-3.2)-(-1.2) = -2

*a*_{3} – *a*_{2} = (-5.2)-(-3.2) = -2

*a*_{4} – *a*_{3} = (-7.2)-(-5.2) = -2

here common difference is same every time.

Since , *d* = -2 and the given series are in A.P.

Hence, next three terms ;

*a*_{5} = – 7.2-2 = -9.2

*a*_{6} = – 9.2-2 = – 11.2

*a*_{7} = – 11.2-2 = – 13.2

(iv) Given, -10, – 6, – 2, 2 …

Here, the common difference are;

*a*_{2} – *a*_{1} = (-6)-(-10) = 4

*a*_{3} – *a*_{2} = (-2)-(-6) = 4

*a*_{4} – *a*_{3} = (2 -(-2) = 4

the common difference is same every time.

Therefore, *d* = 4 and the given numbers are in A.P.

Hence, next three terms are;

*a*_{5} = 2+4 = 6

*a*_{6} = 6+4 = 10

*a*_{7} = 10+4 = 14

(v) Given, 3, 3+√2, 3+2√2, 3+3√2

Here,

*a*_{2} – *a*_{1} = 3+√2-3 = √2

*a*_{3} – *a*_{2} = (3+2√2)-(3+√2) = √2

*a*_{4} – *a*_{3} = (3+3√2) – (3+2√2) = √2

the common difference is same every time.

Therefore, *d* = √2 and the given series forms a A.P.

Hence, next three terms

*a*_{5} = (3+√2) +√2 = 3+4√2

*a*_{6} = (3+4√2)+√2 = 3+5√2

*a*_{7} = (3+5√2)+√2 = 3+6√2

(vi) 0.2, 0.22, 0.222, 0.2222 ….

Here,

*a*_{2} – *a*_{1} = 0.22-0.2 = 0.02

*a*_{3} – *a*_{2} = 0.222-0.22 = 0.002

*a*_{4} – *a*_{3} = 0.2222-0.222 = 0.0002

the common difference is not same every time. and the given series doesn’t forms a A.P.

(vii) 0, -4, -8, -12 …

Here,

*a*_{2} – *a*_{1} = (-4)-0 = -4

*a*_{3} – *a*_{2} = (-8)-(-4) = -4

*a*_{4} – *a*_{3} = (-12)-(-8) = -4

the common difference is same every time.

So, *d* = -4 and the given series forms a A.P.

Hence, next three terms

*a*_{5} = -12-4 = -16

*a*_{6} = -16-4 = -20

*a*_{7} = -20-4 = -24

(viii) -1/2, -1/2, -1/2, -1/2 ….

Here,

*a*_{2} – *a*_{1} = (-1/2) – (-1/2) = 0

*a*_{3} – *a*_{2} = (-1/2) – (-1/2) = 0

*a*_{4} – *a*_{3} = (-1/2) – (-1/2) = 0

the common difference is same every time.

Therefore, *d* = 0 and the given series forms a A.P.

Hence, next three terms

*a*_{5} = (-1/2)-0 = -1/2

*a*_{6} = (-1/2)-0 = -1/2

*a*_{7} = (-1/2)-0 = -1/2

(ix) 1, 3, 9, 27 …

Here,

*a*_{2} – *a*_{1} = 3-1 = 2

*a*_{3} – *a*_{2} = 9-3 = 6

*a*_{4} – *a*_{3} = 27-9 = 18

the common difference is not same every time.

Therefore, and the given series doesn’t form a A.P.

(x) *a*, 2*a*, 3*a*, 4*a* …

Here,

*a*_{2} – *a*_{1} = 2*a*–*a *= *a*

*a*_{3} – *a*_{2} = 3*a*-2*a* = *a*

*a*_{4} – *a*_{3} = 4*a*-3*a* = *a*

the common difference is same every time.

Therefore, *d* = *a* and the given series forms a A.P.

Hence, next three terms

*a*_{5} = 4*a*+*a* = 5*a*

*a*_{6} = 5*a*+*a* = 6*a*

*a*_{7} = 6*a*+*a* = 7*a*

(xi) *a*, *a*^{2}, *a*^{3}, *a*^{4} …

Here,

*a*_{2} – *a*_{1} = *a*^{2}–*a* = a(*a*-1)

*a*_{3} – *a*_{2} = *a*^{3 }–^{ }*a*^{2
}= *a*^{2}(*a*-1)

*a*_{4} – *a*_{3} = *a*^{4} – *a*^{3 }=
*a*^{3}(*a*-1)

the common difference is not same every time.

Therefore, the given series doesn’t forms a A.P.

(xii) √2, √8, √18, √32 …

Here,

*a*_{2} – *a*_{1} = √8-√2 = 2√2-√2 = √2

*a*_{3} – *a*_{2} = √18-√8 = 3√2-2√2 = √2

*a*_{4} – *a*_{3} = 4√2-3√2 = √2

the common difference is same every time.

Therefore, *d* = √2 and the given series forms a A.P.

Hence, next three terms are;

*a*_{5} = √32+√2 = 4√2+√2 = 5√2 = √50

*a*_{6} = 5√2+√2 = 6√2 = √72

*a*_{7} = 6√2+√2 = 7√2 = √98

(xiii) √3, √6, √9, √12 …

Here,

*a*_{2} – *a*_{1} = √6-√3 = √3×√2-√3 = √3(√2-1)

*a*_{3} – *a*_{2} = √9-√6 = 3-√6 = √3(√3-√2)

*a*_{4} – *a*_{3} = √12 – √9 = 2√3 – √3×√3 = √3(2-√3)

the common difference is not same every time.

Therefore, the given series doesn’t form a A.P.

(xiv) 1^{2}, 3^{2}, 5^{2}, 7^{2} …

Or, 1, 9, 25, 49 …..

Here,

*a*_{2} − *a*_{1} = 9−1 = 8

*a*_{3} − *a*_{2 }= 25−9 = 16

*a*_{4} − *a*_{3} = 49−25 = 24

the common difference is not same every time.

Therefore, the given series doesn’t form a A.P.

(xv) 1^{2}, 5^{2}, 7^{2}, 73 …

Or 1, 25, 49, 73 …

Here,

*a*_{2} − *a*_{1} = 25−1 = 24

*a*_{3} − *a*_{2 }= 49−25 = 24

*a*_{4} − *a*_{3} = 73−49 = 24

the common difference is same every time.

Therefore, *d* = 24 and the given series forms a A.P.

Hence, next three terms

*a*_{5} = 73+24 = 97

*a*_{6} = 97+24 = 121

*a*_{7 }= 121+24 = 145