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NCERT Solutions for Class 10th Maths: Chapter 5–Arithematic Progressions



Question 1
In which of the following situations, does the list of numbers involved make an arithmetic Progression, and why?
  1. The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.
  2. The amount of air present in a cylinder when a vacuum pump removes ¼  of the air remaining in the cylinder at a time.
  3. 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.
  4. The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8 % per annum
Solution:
  1. 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
  2.  Let a be the amount of air initially Amount of air remaining after 1st pump =a –(a/4)=3a/4
    Amount of air remaining after 2nd  pump= (3a/4) – (1/4)(3a/4)=9a/16
    Amount of air remaining after 3rd  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
  3. 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
  4.     According to the question Money in account initially=10000
    Money after 1st year  =10000(1+.08)
    Money after 2nd year=10000(1+.08)(1+.08)
    So the series is
    10000,10800,11664….
    Difference between 2nd and Ist term=800
    Difference between 3rd and 2nd term=864
    As difference is not constant, it is not a AP
Question 2
Write first four terms of the AP, when the first term a and the common difference d are given as follows:
  1. a = 10, d = 10
  2.  a = –2, d = 0
  3. a = 4, d = – 3
  4. a=-1 ,d=1/2
  5. a = – 1.25, d = – 0.25
Solution:
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
  1.  a=10,d=20 Series is  10,30,50,70
  2. a=-2  ,d=0 Series is  -2,-2,-2,-2
  3. a=4,d=-3 Series is  4,1,-2,-5
  4. a=-1,d=1/2 series is -1,-1/2,0,1/2
  5. a=-1.25,d=-.25 series is -1.25,-1.50,-1.75,-2
Question 3
For the following APs, write the first term and the common difference:
  1. 3, 1, – 1, – 3, . . .
  2. – 5, – 1, 3, 7, . . .
  3. 1/3 , 5/3 , 9/3 , 13/3 ,…..
  4. 0.6, 1.7, 2.8, 3.9, . . .
Solution:
For any AP, First term is the number in the series  and common difference is defined as difference of second term and first term
  1. 3, 1, – 1, – 3, . . . First term=3
    Common difference=1-3=-2
  2. – 5, – 1, 3, 7, . . . First term=-5
    Common difference=-1-(-5)=4
  3. 1/3 , 5/3 , 9/3 , 13/3 ,….. First term=1/3
    Common difference=(5/3)-(1/3)=4/3
  4. 0.6, 1.7, 2.8, 3.9, . . First term=.6
    Common difference=1.7-.6=1.1
Question 4 :- Which of the following are APs? If they form an A.P. find the common difference d and write three more terms.
(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, a2, a3, a4
(xii) $\sqrt 2 ,\sqrt 8 ,\sqrt {18} ,\sqrt {32} ,.......$
(xiii) $\sqrt 3 ,\sqrt 6 ,\sqrt 9 ,\sqrt {12} ,.......$
(xiv) 12, 32, 52, 72
(xv) 12, 52, 72, 73 …
Solution:-
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
a1 , a2 , a3 ,a4
For the series to be AP,below should be true
    d=  a2 –a1 = a3 –a2 = a4 –a3          ………….(1)
If the series is AP ,then next term would
a5=a4 +d
a6=a4 +2d
a7=a4 +3d
Now let us solves all the section as per theory given above
(i) Here we have,
2, 4, 8, 16 … ……

So putting the values of a1 , a2 , a3 ,a4 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 a1 , a2 , a3 ,a4 equation 1 we find
D= 1/2=1/2 =1/2
Which is True, So it is AP
Next terms of AP are
a5=a4 +d = 7/2 + ½=4
a6=a4 +2d = 7/2 +1=9/2
a7=a4 +3d=5

(iii) Here we have,
− 1.2, − 3.2, − 5.2, − 7.2 …

So putting the values of a1 , a2 , a3 ,a4 equation 1 we find $D= -2=-2=-2 $
Which is true,So it is AP
Next terms of AP are
a5=a4 + d = -7.2 + (-2) = -9.2
a6=a4 + 2d = -7.2 + 2(-2) = -11.2
a7=a4 + 3d = -7.2 + 3(-2) = -13.2

(iv) Here we have,
− 10, − 6, − 2, 2 … … …

So putting the values of a1 , a2 , a3 ,a4 equation 1 we find $d= 4=4=4 $

Which is true,So it is AP
Next terms of AP are
a5=a4 + d = 2 + (4) = 6
a6=a4 + 2d = 2 + 2(4) = 10
a7=a4 + 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 a1 , a2 , a3 ,a4 equation 1 we find $D = \sqrt 2 = \sqrt 2 = \sqrt 2 $
Which is true ,So it is AP
Next terms of AP are
a5=a4 + d = $3 + 4\sqrt 2 $
a6=a4 + 2d = $3 + 5\sqrt 2 $
a7=a4 + 3d = $3 + 6\sqrt 2 $

(vi) Here we have,
0.2, 0.22, 0.222, 0.2222 … ………
So putting the values of a1 , a2 , a3 ,a4 equation 1 we find
$D = .02 = .002 = .0003$
Clearly this is not true,So it is not AP

(vii) Here we have,
0, − 4, − 8, − 12 …
So putting the values of a1 , a2 , a3 ,a4 equation 1 we find
$D=-4=-4=-4$
Which is true ,So it is AP
Next terms of AP are
a5=a4 + d = -12 + (-4) = -16
a6=a4 + 2d = -12 + 2(-4) = -20
a7=a4 + 3d = -12 + 3(-4) = -24

(viii) Here we have,
-1/2, -1/2,-1/2,-1/2…………
So putting the values of a1 , a2 , a3 ,a4 equation 1 we find
$D=0=0=0$
So it is AP with zero Common difference
Next terms of AP are
a5=a4 + d = $ - {1 \over 2}$
a6=a4 + 2d = $ - {1 \over 2}$
a7=a4 + 3d = $ - {1 \over 2}$

(ix) Here we have,
1, 3, 9, 27 …………
So putting the values of a1 , a2 , a3 ,a4 equation 1 we find
$D=2=6=18$
Clearly not an AP

(x) Here we have,
a, 2a, 3a, 4a …………
So putting the values of a1 , a2 , a3 ,a4 equation 1 we find
D=a=a=a
Clearly an AP
Next terms of AP are
a5=a4 + d = 5a
a6=a4 + 2d = 6a
a7=a4 + 3d = 7a

(xi) Here we have,
a, a2, a3, a4 …………
So putting the values of a1 , a2 , a3 ,a4 equation 1 we find
$D = {a^2}-a = {a^3} - {\rm{ }}{a^2}_ = {a^4}-{\rm{ }}{a^3}$
Clearly not an AP

(xii) Here we have,
$\sqrt 2 ,\sqrt 8 ,\sqrt {18} ,\sqrt {32} ,.......$
It can rewritten as
$\sqrt 2 ,2\sqrt 2 ,3\sqrt 2 4\sqrt 2 .......$
So putting the values of a1 , a2 , a3 ,a4 equation 1 we find
$D = \sqrt 2 = \sqrt 2 = \sqrt 2 $
Which is true ,So it is AP
Next terms of AP are
a5=a4 + d = $5\sqrt 2 $
a6=a4 + 2d = $6\sqrt 2 $
a7=a4 + 3d = $7\sqrt 2 $

(xiii) Here we have,
$\sqrt 3 ,\sqrt 6 ,\sqrt 9 ,\sqrt {12} ,.......$
So putting the values of a1 , a2 , a3 ,a4 equation 1 we find
$d = \sqrt 6 - \sqrt 3 = \sqrt 9 - \sqrt 6 = \sqrt {12} - \sqrt 9 $
Clearly not a AP

(xiv) Here we have,
12, 32, 52, 72 …………
So putting the values of a1 , a2 , a3 ,a4 equation 1 we find
D=8=16=24
Not an AP

(xv) Here we have,
12, 52, 72, 73 …………
So putting the values of a1 , a2 , a3 ,a4 equation 1 we find
D=24=24=24
Which is true ,So it is AP
Next terms of AP are
a5=a4 + d = 97
a6=a4 + 2d = 121
a7=a4 + 3d = 145


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