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, 3V/4 , (3V/4)2 , (3V/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 a1, a2, a3, a4, a5 …
a1 = a = 10
a2 = a1+d = 10+10 = 20
a3 = a2+d = 20+10 = 30
a4 = a3+d = 30+10 = 40
a5 = a4+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 a1, a2, a3, a4, a5 …
a1 = a = -2
a2 = a1+d = – 2+0 = – 2
a3 = a2+d = – 2+0 = – 2
a4 = a3+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 a1, a2, a3, a4, a5 …
a1 = a = 4
a2 = a1+d = 4-3 = 1
a3 = a2+d = 1-3 = – 2
a4 = a3+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 a1, a2, a3, a4, a5 …
a2 = a1+d = -1+1/2 = -1/2
a3 = a2+d = -1/2+1/2 = 0
a4 = a3+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 a1, a2, a3, a4, a5 …
a1 = a = – 1.25
a2 = a1 + d = – 1.25-0.25 = – 1.50
a3 = a2 + d = – 1.50-0.25 = – 1.75
a4 = a3 + 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.
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, 2a, 3a, 4a …
(xi) a, a2, a3, a4 …
(xii) √2, √8, √18, √32 …
(xiii) √3, √6, √9, √12 …
(xiv) 12, 32, 52, 72 …
(xv) 12, 52, 72, 73 …
(i) here ,
2, 4, 8, 16 …
Here, the common difference is;
a2 – a1 = 4 – 2 = 2
a3 – a2 = 8 – 4 = 4
a4 – a3 = 16 – 8 = 8
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,
a2 – a1 = 5/2-2 = 1/2
a3 – a2 = 3-5/2 = 1/2
a4 – a3 = 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
a5 = 7/2+1/2 = 4
a6 = 4 +1/2 = 9/2
a7 = 9/2 +1/2 = 5
(iii) Given, -1.2, – 3.2, -5.2, -7.2 …
Here,
a2 – a1 = (-3.2)-(-1.2) = -2
a3 – a2 = (-5.2)-(-3.2) = -2
a4 – a3 = (-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 ;
a5 = – 7.2-2 = -9.2
a6 = – 9.2-2 = – 11.2
a7 = – 11.2-2 = – 13.2
(iv) Given, -10, – 6, – 2, 2 …
Here, the common difference are;
a2 – a1 = (-6)-(-10) = 4
a3 – a2 = (-2)-(-6) = 4
a4 – a3 = (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;
a5 = 2+4 = 6
a6 = 6+4 = 10
a7 = 10+4 = 14
(v) Given, 3, 3+√2, 3+2√2, 3+3√2
Here,
a2 – a1 = 3+√2-3 = √2
a3 – a2 = (3+2√2)-(3+√2) = √2
a4 – a3 = (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
a5 = (3+√2) +√2 = 3+4√2
a6 = (3+4√2)+√2 = 3+5√2
a7 = (3+5√2)+√2 = 3+6√2
(vi) 0.2, 0.22, 0.222, 0.2222 ….
Here,
a2 – a1 = 0.22-0.2 = 0.02
a3 – a2 = 0.222-0.22 = 0.002
a4 – a3 = 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,
a2 – a1 = (-4)-0 = -4
a3 – a2 = (-8)-(-4) = -4
a4 – a3 = (-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
a5 = -12-4 = -16
a6 = -16-4 = -20
a7 = -20-4 = -24
(viii) -1/2, -1/2, -1/2, -1/2 ….
Here,
a2 – a1 = (-1/2) – (-1/2) = 0
a3 – a2 = (-1/2) – (-1/2) = 0
a4 – a3 = (-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
a5 = (-1/2)-0 = -1/2
a6 = (-1/2)-0 = -1/2
a7 = (-1/2)-0 = -1/2
(ix) 1, 3, 9, 27 …
Here,
a2 – a1 = 3-1 = 2
a3 – a2 = 9-3 = 6
a4 – a3 = 27-9 = 18
the common difference is not same every time.
Therefore, and the given series doesn’t form a A.P.
(x) a, 2a, 3a, 4a …
Here,
a2 – a1 = 2a–a = a
a3 – a2 = 3a-2a = a
a4 – a3 = 4a-3a = a
the common difference is same every time.
Therefore, d = a and the given series forms a A.P.
Hence, next three terms
a5 = 4a+a = 5a
a6 = 5a+a = 6a
a7 = 6a+a = 7a
(xi) a, a2, a3, a4 …
Here,
a2 – a1 = a2–a = a(a-1)
a3 – a2 = a3 – a2 = a2(a-1)
a4 – a3 = a4 – a3 = a3(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,
a2 – a1 = √8-√2 = 2√2-√2 = √2
a3 – a2 = √18-√8 = 3√2-2√2 = √2
a4 – a3 = 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;
a5 = √32+√2 = 4√2+√2 = 5√2 = √50
a6 = 5√2+√2 = 6√2 = √72
a7 = 6√2+√2 = 7√2 = √98
(xiii) √3, √6, √9, √12 …
Here,
a2 – a1 = √6-√3 = √3×√2-√3 = √3(√2-1)
a3 – a2 = √9-√6 = 3-√6 = √3(√3-√2)
a4 – a3 = √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) 12, 32, 52, 72 …
Or, 1, 9, 25, 49 …..
Here,
a2 − a1 = 9−1 = 8
a3 − a2 = 25−9 = 16
a4 − a3 = 49−25 = 24
the common difference is not same every time.
Therefore, the given series doesn’t form a A.P.
(xv) 12, 52, 72, 73 …
Or 1, 25, 49, 73 …
Here,
a2 − a1 = 25−1 = 24
a3 − a2 = 49−25 = 24
a4 − a3 = 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
a5 = 73+24 = 97
a6 = 97+24 = 121
a7 = 121+24 = 145