Arithematic progression

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 Rs 15 for the first km and Rs 8 for each additional km
Solution:The fare for first km = 15
the taxi fare after completing 1st km=8

always, fare=(number of km travelled)×(cost for each km),

Therefore,
fare for 1st km=15              =15
fare for 2nd km=15+8        =23
fare for 3rd km=15+(2×8)  =31
fare for 4th km=15+(3×8)   =39

a1=15
a2=23
a3=31
a4=39

d1=a2-a1=23-15=8
d2=a3-a2=31-23=8
d3=a4-a3=39-31=8

since d1=d2=d3
there above sequence is in ap
               
The same problem is solved in Youtube
https://youtu.be/oXGPpZXDxZI

The same problem is solved handy or with hand in the following link
https://mytextbookexercise.blogspot.com/2019/05/exercise-5.html


(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.
Solution: let the amount of air present in the cylinder be 'V'
Everytime 1/4 th of air goes,

Therefore
initially air was='V'
after 1/4th of air goes=V - (1/4)V  (take l.c.m)
                                     =3V/4
again after 1/4th of the air goes
                          =  3V/4 - (1/4)×3V/4
                             =3V/4 - 3V/16
                             =(12V-3V)/16
                             =9V/16         
Thus,
a1= V
a2=3V/4
a3=9V/16

d1=a2 - a1 = 3V/4 - V = -V/4
d2=a3 - a2 = 9V/4 - 3V/4 = 6V/4

since d1 not equal to d2
therefore above sequence is not in ap     
               
                           

(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.
Solution: 
cost for 1st meter= 150
cost for 2nd meter= 150+50= 200
cost for 3rd meer= 150+(2×50)
                               = 150+100= 250
cost for 4th meter= 150+(3×50)
                                   150+150=300

Therefore
a1= 150
a2= 200
a3= 250
a4= 300

thus,
d1=a2 - a1=200 - 150=50
d2=a3 - a2=250 - 200=50
d3=a4 - a3=300 - 250=50

since d1=d2=d3, therefore the above sequence is in ap.

(iv) The amount of money in the account every year, when Rs 10000 is deposited at
compound interest at 8 % per annum.
Solution:
Here the principal 'p'= 10,000
                                   r = 8
                                  n= 1st year or 2nd yr etc

Formula for compound intrest is
                            =p[1+(r/100)]^n
          '^' means 'power of'
                         
compound intrest for 1st year
                            =10,000[1+(8/100)]^1
                            =10,000[1+8/100] (take l.c.m)
                           =10,000[(108/100]
                           =10,0×108 = 10,800

compound intrest for 2nd year
                          = 10,000[1+(8/100)]^2
                          =10,000[108/100]^2
                          =10,000×(108/100)×(108/100)
                          =11664

compound intrest for 3rd year
                        =10,000[1+(8/100)]^3
                        =10,000[108/100]^3
                        =10,000×(108/100)×(108/100)
                                       ×(108/100)
                       =12597.12

compound intrest for 4th year
                       =10,000[1+(8/100)]^4
                       =10,000[108/100]^4
          =10,000×(108/100)×(108/100)
                        ×(108/100)×(108/100)
         =13,604.8896

Therefore
a1=10,800
a2=11664
a3=12597.12
a4=13,604.8896

d1= a2 - a1 = 11664 - 10800 =844
d2= a3 - a2 = 12597.12 - 11664=933.12
d3= a4 - a3 = 13604.8896 - 12597.12=1007.76

since d1 is not equal to d2
   and d2 is not equal to d3

therefore the above sequence is not in 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
Solution:
a1=10
a2=a1+d=10+10=20
a3=a2+d=20+10=30
a4=a3+d=30+10=40

therefore the first four terms are 10,20,30,40


(ii) a = –2, d = 0
Solution:
a1=(-2)
a2= a1+d=[(-2)+0]=  -2
a3= a2+d=[(-2)+(-2)]=  -4
a4= a3+d= [(-4)+(-2)]= -6

therefore tge first four terms are
-2, -2, -4, -6

(iii) a = 4, d = – 3
Solution:
a1=4
a2=a1+d=4+(-3)=4-3=1
a3=a2+d=1+(-3)=1-3=  -2
a4=a3+d=[(-2)+(-3)]=  -2-3=  -5

therefore the first four terms are
4,1, -2,  -5

(iv) a = – 1, d =1/2
Solution:
a1=  -1
a2= a1+d =  -1+(1/2)  =  -(1/2)
a3= a2+d=  [-(1/2)+(1/2)] = 0
a4= a3+d= [0+(1/2)]= 1/2

 therefore the first four terms are
-1, -(1/2),  0,  1/2


(v) a = – 1.25, d = – 0.25
Solution:

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

(i) 3, 1, – 1, – 3, . . .
Solution:

(ii) – 5, – 1, 3, 7, . . .
Solution:

(iii)
1/3, 5/3, 9/3, 1/33 . . .
Solution:

(iv) 0.6, 1.7, 2.8, 3.9, . . .
Solution:

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, . . .
Solution:

 (ii) 2, 5/2, 3, 7/2,...
Solution:

(iii) – 1.2, – 3.2, – 5.2, – 7.2, . . .
Solution:

 (iv) – 10, – 6, – 2, 2, . . .
Solution:

(v) 3, 3+root2  , 3+2root2  , 3+3root2,  . . .
Solution:

 (vi) 0.2, 0.22, 0.222, 0.2222, . . .
Solution:

(vii) 0, – 4, – 8, –12, . . .
Solutions:

(viii) –1/2, –1/2, –1/2, –1/2
Solutions:

ix) 1, 3, 9, 27, . . .
Solutions:

 (x) a, 2a, 3a, 4a, . . .
Solutions:

(xi) a, a2, a3, a4, . . . 
Solutions:

(xii) 2, 8, 18 , 32, . . .
Solutions:

(xiii) 3, 6, 9 , 12 , . . .
Solutions:

 (xiv) 12, 32, 52, 72, . . .
Solutions:

(xv) 12, 52, 72, 73, . . 
Solutions: