NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.1
In this page we have NCERT Solutions for Class 10th Maths: Chapter 5 - Arithmetic Progressions for
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An arithmetic progression (AP) is a list of numbers in which each term is obtained by adding a fixed number d to the preceding term, except the first term. The fixed number d is called the common difference.
The general form of an AP is $a$, $a + d$, $a + 2d$, $a + 3d$,.....
A given list of numbers $a_1$, $a_2$, $a_3$, .... is an AP, if
$d= a_2 - a_1= a_3 - a_2 =a_4 - a_3, ..... ,a_{k + 1} – a_k$
Question 1
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
Solution:
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
Question 2
Write first four terms of the AP, when the first term a and the common difference d are given as follows:
a = 10, d = 10
a = –2, d = 0
a = 4, d = – 3
a=-1 ,d=1/2
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
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
Question 3
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, . . .
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
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
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, a^{2}, a^{3}, a^{4} …
(xii) $\sqrt 2 ,\sqrt 8 ,\sqrt {18} ,\sqrt {32} ,.......$
(xiii) $\sqrt 3 ,\sqrt 6 ,\sqrt 9 ,\sqrt {12} ,.......$
(xiv) 1^{2}, 3^{2}, 5^{2}, 7^{2} …
(xv) 1^{2}, 5^{2}, 7^{2}, 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
a_{1} , a_{2} , a_{3} ,a_{4}
For the series to be AP,below should be true
d= a_{2} –a_{1} = a_{3} –a_{2} = a_{4} –a_{3 } ………….(1)
If the series is AP ,then next term would
a_{5}=a_{4} +d
a_{6}=a_{4} +2d
a_{7}=a_{4} +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 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
(vii)
Here we have,
0, − 4, − 8, − 12 …
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 = -12 + (-4) = -16
a_{6}=a_{4} + 2d = -12 + 2(-4) = -20
a_{7}=a_{4} + 3d = -12 + 3(-4) = -24
(viii)
Here we have,
-1/2, -1/2,-1/2,-1/2…………
So putting the values of a_{1} , a_{2} , a_{3} ,a_{4} equation 1 we find
$D=0=0=0$
So it is AP with zero Common difference
Next terms of AP are
a_{5}=a_{4} + d = $ - {1 \over 2}$
a_{6}=a_{4} + 2d = $ - {1 \over 2}$
a_{7}=a_{4} + 3d = $ - {1 \over 2}$
(ix)
Here we have,
1, 3, 9, 27 …………
So putting the values of a_{1} , a_{2} , a_{3} ,a_{4} 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 a_{1} , a_{2} , a_{3} ,a_{4} equation 1 we find
D=a=a=a
Clearly an AP
Next terms of AP are
a_{5}=a_{4} + d = 5a
a_{6}=a_{4} + 2d = 6a
a_{7}=a_{4} + 3d = 7a
(xi)
Here we have,
a, a^{2}, a^{3}, a^{4} …………
So putting the values of a_{1} , a_{2} , a_{3} ,a_{4} 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 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 = $5\sqrt 2 $
a_{6}=a_{4} + 2d = $6\sqrt 2 $
a_{7}=a_{4} + 3d = $7\sqrt 2 $
(xiii)
Here we have,
$\sqrt 3 ,\sqrt 6 ,\sqrt 9 ,\sqrt {12} ,.......$
So putting the values of a_{1} , a_{2} , a_{3} ,a_{4} 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,
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=8=16=24
Not an AP
(xv)
Here we have,
1^{2}, 5^{2}, 7^{2}, 73 …………
So putting the values of a_{1} , a_{2} , a_{3} ,a_{4} equation 1 we find
D=24=24=24
Which is true ,So it is AP
Next terms of AP are
a_{5}=a_{4} + d = 97
a_{6}=a_{4} + 2d = 121
a_{7}=a_{4} + 3d = 145
Summary
NCERT Solutions for Class 10th Maths: Chapter 5 - Arithmetic Progressions Ex 5.1 has been prepared by Expert with utmost care. If you find any mistake.Please do provide feedback on mail.You can download this as pdf Download this assignment as pdf
This chapter 5 has total 4 Exercise 5.1 ,5.2,5.3 and 5.4. This is the First exercise in the chapter.You can explore previous exercise of this chapter by clicking the link below