- Introduction
- |
- What is arithmetic progression
- |
- Some Important points about AP
- |
- nth term of Arithmetic Progression
- |
- Sum of nth item in Arithmetic Progression

- Introduction
- |
- What is arithmetic progression
- |
- Some Important points about AP
- |
- nth term of Arithmetic Progression
- |
- Sum of nth item in Arithmetic Progression

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

- The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.
- The amount of air present in a cylinder when a vacuum pump removes ¼ of the air remaining in the cylinder at a time.
- The cost of digging a well after every meter of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent meter.
- The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8 % per annum

- According to the question
Fare for First km= Rs 15

Fare for first km + additional 1 km = 15 +8

Fare for first km + additional 2 km = 15 +2X8

Fare for first km + additional 3 km = 15 +3X8

So series is like

15, 15 +8, 15 +2X8, 15 +3X8 ………..

Difference between two terms =8 everywhere except first term

So it is Arithmetic Progression

- Let a be the amount of air initially
Amount of air remaining after 1
^{st}pump =a –(a/4)=3a/4

Amount of air remaining after 2^{nd}pump= (3a/4) – (1/4)(3a/4)=9a/16

Amount of air remaining after 3^{rd}pump= (9a/16) – (1/4)(9a/16)=27a/64

So the series is like

a,3a/4,9a/4,27a/64……..

Difference Ist and second term=-a/4

Difference between Second and Third term= -3a/16

So difference is not constant

So it is not Arithmetic Progression

- According to the question
Cost of digging for First m= Rs 150

Cost of digging for First m + additional 1 m = 150 +50

Cost of digging for First m + additional 2 m = 150+2X50

Cost of digging for First m + additional 3 m = 150 +3X50

So series is like

150, 150 +50, 150 +2X50, 150 +3X50 ………..

Difference between two terms =50 everywhere except first term

So it is Arithmetic Progression

- According to the question
Money in account initially=10000

Money after 1^{st}year =10000(1+.08)

Money after 2^{nd}year=10000(1+.08)(1+.08)

So the series is

10000,10800,11664….

Difference between 2^{nd}and Ist term=800

Difference between 3^{rd}and 2^{nd}term=864

As difference is not constant, it is not a AP

Write first four terms of the AP, when the first term

*a*= 10,*d*= 10-
*a*= –2,*d*= 0 *a*= 4,*d*= – 3*a=-1 ,d=1/2**a*= – 1.25,*d*= – 0.25

Arithmetic Progression with first term a and common difference is shown

a,a+d,a+2d,a+3d……

Solving all these questions on the based of these formula

- a=10,d=20 Series is 10,30,50,70
- a=-2 ,d=0
Series is -2,-2,-2,-2

- a=4,d=-3
Series is 4,1,-2,-5
- a=-1,d=1/2
series is -1,-1/2,0,1/2

- a=-1.25,d=-.25
series is -1.25,-1.50,-1.75,-2

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

- 3, 1, – 1, – 3, . . .

- – 5, – 1, 3, 7, . . .

- 1/3 , 5/3 , 9/3 , 13/3 ,…..

- 0.6, 1.7, 2.8, 3.9, . . .

For any AP, First term is the number in the series and common difference is defined as difference of second term and first term

- 3, 1, – 1, – 3, . . .
First term=3

Common difference=1-3=-2

- – 5, – 1, 3, 7, . . .
First term=-5

Common difference=-1-(-5)=4

- 1/3 , 5/3 , 9/3 , 13/3 ,…..
First term=1/3

Common difference=(5/3)-(1/3)=4/3

- 0.6, 1.7, 2.8, 3.9, . .
First term=.6

Common difference=1.7-.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 + \sqrt {2}, 3 + 2\sqrt {2}, 3 + 3\sqrt {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, a

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

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

(xiv) 1

(xv) 1

For Arithmetic Progression, Common Difference should be same across

$a,a+d,a+2d,a+3d$

Lets us assume four term given of series as

a

For the series to be AP,below should be true

d= a

If the series is AP ,then next term would

a

a

a

Now let us solves all the section as per theory given above

2, 4, 8, 16 … ……

So putting the values of a_{1} , a_{2} , a_{3} ,a_{4} equation 1 we find $D=2=4=8 $

Which is not true, So it is not AP

**(ii)**
Here we have,

2, 5/2,3,7/2 … … … …

So putting the values of a_{1} , a_{2} , a_{3} ,a_{4} equation 1 we find

D= 1/2=1/2 =1/2

Which is True, So it is AP

Next terms of AP are

a_{5}=a_{4} +d = 7/2 + ½=4

a_{6}=a_{4} +2d = 7/2 +1=9/2

a_{7}=a_{4} +3d=5

**(iii)**
Here we have,

− 1.2, − 3.2, − 5.2, − 7.2 …

So putting the values of a_{1} , a_{2} , a_{3} ,a_{4} equation 1 we find $D= -2=-2=-2 $

Which is true,So it is AP

Next terms of AP are

a_{5}=a_{4} + d = -7.2 + (-2) = -9.2

a_{6}=a_{4} + 2d = -7.2 + 2(-2) = -11.2

a_{7}=a_{4} + 3d = -7.2 + 3(-2) = -13.2

**(iv)**
Here we have,

− 10, − 6, − 2, 2 … … …

So putting the values of a_{1} , a_{2} , a_{3} ,a_{4} equation 1 we find $d= 4=4=4 $

Which is true,So it is AP

Next terms of AP are

a_{5}=a_{4} + d = 2 + (4) = 6

a_{6}=a_{4} + 2d = 2 + 2(4) = 10

a_{7}=a_{4} + 3d = 2 + 3(4) = 14

**(v)**
Here we have,

$3, 3 + \sqrt {2}, 3 + 2\sqrt {2}, 3 + 3\sqrt {2}...... $

So putting the values of a_{1} , a_{2} , a_{3} ,a_{4} equation 1 we find $D = \sqrt 2 = \sqrt 2 = \sqrt 2 $

Which is true ,So it is AP

Next terms of AP are

a_{5}=a_{4} + d = $3 + 4\sqrt 2 $

a_{6}=a_{4} + 2d = $3 + 5\sqrt 2 $

a_{7}=a_{4} + 3d = $3 + 6\sqrt 2 $

**(vi)**
Here we have,

0.2, 0.22, 0.222, 0.2222 … ………

So putting the values of a_{1} , a_{2} , a_{3} ,a_{4} equation 1 we find

$D = .02 = .002 = .0003$

Clearly this is not true,So it is not AP

0, − 4, − 8, − 12 …

So putting the values of a

$D=-4=-4=-4$

Which is true ,So it is AP

Next terms of AP are

a

a

a

-1/2, -1/2,-1/2,-1/2…………

So putting the values of a

$D=0=0=0$

So it is AP with zero Common difference

Next terms of AP are

a

a

a

1, 3, 9, 27 …………

So putting the values of a

$D=2=6=18$

Clearly not an AP

a, 2a, 3a, 4a …………

So putting the values of a

D=a=a=a

Clearly an AP

Next terms of AP are

a

a

a

a, a

So putting the values of a

$D = {a^2}-a = {a^3} - {\rm{ }}{a^2}_ = {a^4}-{\rm{ }}{a^3}$

Clearly not an AP

$\sqrt 2 ,\sqrt 8 ,\sqrt {18} ,\sqrt {32} ,.......$

It can rewritten as

$\sqrt 2 ,2\sqrt 2 ,3\sqrt 2 4\sqrt 2 .......$

So putting the values of a

$D = \sqrt 2 = \sqrt 2 = \sqrt 2 $

Which is true ,So it is AP

Next terms of AP are

a

a

a

$\sqrt 3 ,\sqrt 6 ,\sqrt 9 ,\sqrt {12} ,.......$

So putting the values of a

$d = \sqrt 6 - \sqrt 3 = \sqrt 9 - \sqrt 6 = \sqrt {12} - \sqrt 9 $

Clearly not a AP

1

So putting the values of a

D=8=16=24

Not an AP

1

So putting the values of a

D=24=24=24

Which is true ,So it is AP

Next terms of AP are

a

a

a

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