Arithmetic Progression class-10th math Exercise 5.1 solution
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:- First taxi fare for one km is RS 15
2nd taxi fare for km is RS(15+8)=RS 23
3rd taxi fare for km is RS(23+8)=RS 31
4th taxi fare for km is RS(31+8)=RS 39
the Sequence of taxi fare for each km is RS 15,23,31,39......
now common difference(d1)=23-15=8
d2=31-23=8
d3=39-31=8
d1= d2=d3
so, this sequence is aritmetic progression because all of the common difference is Equal.
(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 air present in thye cylinder is =a
again the removing air remains by vaccum pump
=a-a/4
=(4a-a)/4
=3a/4
again the removing air remains by vaccum pump
=3a/4 -(3a/4)(1/4)
=3a/4 -3a/16
= (12a-3a)/16
=9a/16
(इसका मतलब हुआ की जितना हवा सिलिंडर में बचा हुआ है उसका 1/4 भाग हवा बचे हुए हवा मेसे निकाल लेता है।)
again the removing air remains by vaccum pump=9a/16 -(9a/16)(1/4)
=9a/16 -9a/64
=(36a-9a)/64
=27a/64
So on the Sequence 3a/4, 9a/16,27a/16................
d1=2nd term-first term
d1=a2-a1
d1=9a/16-3a/4
d1=-12a/16
d2=a3-a2
d2=27a/16-9a/16
d2=18a/16
No , Because d1≠d2
(iii) The cost of digging a well after every metre of digging, when it costs RS 150 for the first metre and rises by RS50 for each subsequent metre.
Solution:-
The cost of digging for the first metre=150
The cost of digging for the second metre=150+50=200
The cost of digging for the Third metre=200+50=250
so on the Sequence 150,200 ,250................
d1=2nd term-first term
d1=a2-a1
d1=200-150
d1=50
d2=a3-a2
d2=250-200
d2=50
yes this are in AP Because all common difference is Equal.
(iv) The amount of money in the account every year, when RS 10000 is deposited at compound interest at 8 % per annum.
Solution:-
compound intrest =p(1+r/100)^t
p= 10000
r=8 %
t= time
compound intrest for first year = 10000(1+8/100)^1
= 10800
compound intrest for 2nd year = 10000(1+8/100)^2
= 11664
compound intrest for 3rd year = 10000(1+8/100)^3
=12597.12
so on the Sequence 10800,11664 ,12597.12 ................
d1=a2-a1
d1=11664 -10800=
d1=1,664
d2=a3-a2
d2=12597.12-11664
d2=933.12
No, Because d1≠d2
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 (ii) a = –2, d = 0 (iii) a = 4, d = – 3 (iv) a = – 1, d = 1/2 (v) a = – 1.25, d = – 0.2
solution:-
(i) a = 10, d = 10
a1=a+d=10+10=20
a2= a1+d= 20+10=30
a3= a2+d= 30+10=40
a4= a3+d= 40+10=50
first four terms of the AP is 20,30,40,50 Ans
(ii) a = –2, d = 0
a1=a+d=-2+0=-2
a2= a1+d=-2+0=-2
a3= a2+d= -2+0=-2
a4= a3+d= -2+0=-2
first four terms of the AP is -2,-2,-2,-2 ans
(iii) a = 4, d = – 3
a1=a+d=4+(-3)=4-3=1
a2= a1+d=1+(-3)=-2
a3= a2+d= -2+(-3)=-5
a4= a3+d= -5-3=-8
first four terms of the AP is-3,1,-5,-8 ans
(iv) a = – 1, d = 1/2
a1=a+d=-1+1/2=+1/2
a2= a1+d=1/2+1/2=1
a3= a2+d= 1+1/2=3/2
a4= a3+d= 3/2+1/2=2
first four terms of the AP is 1/2,1,3/2,2 ans
(v) a = – 1.25, d = – 0.2
a1=a+d=-1.25+(– 0.2)=-1.27
a2= a1+d=-1.27+(– 0.2)=-1.29
a3= a2+d= -1.29+(– 0.2)=-1.31
a4= a3+d= -1.31+(– 0.2)=-1.33
first four terms of the AP is -1.27,-1.29,-1.31,-1.33 ans
3. For the following APs, write the first term and the common difference:
(i) 3, 1, – 1, – 3, . . .
(ii) – 5, – 1, 3, 7, . . .
(iii) 1/3 , 5/3, 9/3, 13/3 . . .
(iv) 0.6, 1.7, 2.8, 3.9, . . .
solution:-
(i) 3, 1, – 1, – 3, . . .
given that this squences are in AP
the first term(a)=3
the common difference(d)
=a2-a1
=1-3
=-2
hence common difference(d) is -2 ans
(ii) – 5, – 1, 3, 7, . . .
given that this squences are in AP
the first term(a)=-5
the common difference(d)
=a2-a1
=-1-(-5)
=-1+5
=4
hence common difference(d) is 4 ans
(iii) 1/3 , 5/3, 9/3, 13/3 . . .
given that this squences are in AP
the first term(a)=1/3
the common difference(d)
=a2-a1
=5/3 - 1/3
=4/3
hence common difference(d) is 4/3 ans
(iv) 0.6, 1.7, 2.8, 3.9, . . .
given that this squences are in AP
the first term(a)=0.6
the common difference(d)
=a2-a1
=1.7 - 0.6
=1.1
hence common difference(d) is 1.1 ans
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, . . . (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, a^2 , a^3 , a^4 , . . .
(xii) ✓2, ✓8, ✓18 , ✓32, . . .
(xiii) ✓3, ✓6, ✓9 , ✓12 , . . .
(xiv) 1/ 2 , 3/2 , 5/2 , 7/2 , . . .
(xv) 1/2
, 5/2
, 7/2
, 73, . . .
solution:-
(i) 2, 4, 8, 16, . . .
Check this squences are in AP
the first term(a)=2
the common difference(d1)
=a2-a1
=4-2
=2
the common difference(d2)
=a3-a2
= 8-4
=4
No, Because d1≠d2
(ii)) 2,5/2,3,7/2.........
Check this squences are in AP
the first term(a)=2
the common difference(d1)
=a2-a1
=5/2-2
=0.5
the common difference(d2)
=a3-a2
= 3-5/2
=0.5
d1=d2
so the next three more term means that is
a4 =7/2+0.5=4
a5 =4+0.5=4.5
a6 =4.5+0.5=5
three more term is 4 , 4.5 , 5 ans
(iii) – 1.2, – 3.2, – 5.2, – 7.2, . . .
Check this squences are in AP
the first term(a)=-1.2
the common difference(d1)
=a2-a1
=-3.2-(-1.2)
=-3.2+1.2=-2
the common difference(d2)
=a3-a2
= -5.2-(-3.2)
=-5.2+3.2
=-2
d1=d2
so the next three more term means that is
a5 =-7.2+(-2)=-9.2
a6 =-9.2+(-2)=-11.2
a7 =-11.2+(-2)=-13.2
three more term is -9.2 ,-11.2, -13.2 ans
(iv) – 10, – 6, – 2, 2, . . .
Check this squences are in AP
the first term(a)=-10
the common difference(d1)
=a2-a1
=-6-(-10)
=-6+10)=4
the common difference(d2)
=a3-a2
= -2-(-6)
=-2+6
=4
d1=d2so the next three more term means that is
a5 =2+4=6
a6 =6+4=10
a7 =10+4=14
three more term is 6,10,14 ans
(v) 3,3+√2 ,3+2√2 ,3+3√2 . . .
Check this squences are in AP
the first term(a)=3
the common difference(d1)
=a2-a1
=3+√2-3
=√2
the common difference(d2)
=a3-a2
= 3+2√2 -(3+√2)
= 3+2√2 -3-√2
= √2
d1=d2 so the next three more term means that is
a5 =3+3√2+√2=3+4√2
a6 =3+4√2+√2=3+5√2
a7 =3+5√2+√2=3+6√2
three more term is 3+4√2,3+5√2,3+6√2 ans
(vi) 0.2, 0.22, 0.222, 0.2222, . . .
Check this squences are in AP
the first term(a)=0.2
the common difference(d1)
=a2-a1
=0.22- 0.2
=0.02
the common difference(d2)
=a3-a2
= 0.222- 0.22
=0.002
No, Because d1≠d2
(vii) 0, – 4, – 8, –12, . . .
Check this squences are in AP
the first term(a)= 0
the common difference(d1)
=a2-a1
=-4-0
=-4
the common difference(d2)
=a3-a2
= -8-(-4)
=-8+4
=-4
d1=d2so the next three more term means that is
a5 =-12+(-4)=-12-4=-16
a6 =-16+(-4)=-16-4=-20
a7 =-20+(-4)=-20-4=-24
three more term is-16,-20,-24 ans
(viii) – 1/2 , – 1/2 , – 1/2 , – 1/2 , . . .
Check this squences are in AP
the first term(a)= -1/2
the common difference(d1)
=a2-a1
=-1/2-(-1/2)
=-1/2 +1/2 =0
the common difference(d2)
=a3-a2
=-1/2-(-1/2)
=-1/2 +1/2 =0
d1=d2
so the next three more term means that is
a5 =-1/2+0=-1/2
a6 =-1/2+0=-1/2
a7 =-1/2+0=-1/2
three more term is-1/2,-1/2,-1/2 ans
(ix) 1, 3, 9, 27, . . .
Check this squences are in AP
the first term(a)= 1
the common difference(d1)
=a2-a1
=3-1
=2
the common difference(d2)
=a3-a2
=9-3
=6
No, Because d1≠d2
(x) a, 2a, 3a, 4a, . . .
Check this squences are in AP
the first term(a)= -1/2
the common difference(d1)
=a2-a1
=2a-a
=a
the common difference(d2)
=a3-a2
=3a-2a
=a
d1=d2
so the next three more term means that is
a5 =4a+a=5a
a6 =5a+a=6a
a7 =6a+a=7a
three more term is 5a,6a,7a ans
(xi) a, a^2 , a^3 , a^4 , . . .
Check this squences are in AP
the first term(a)= a
the common difference(d1)
=a2-a1
=a^2-a
=a(a-1)
the common difference(d2)
=a3-a2
=a^3-a^2
=a^2(a-1)
No, Because d1≠d2
(xii) ✓2, ✓8, ✓18 , ✓32, . . .
Check this squences are in AP
the first term(a)=✓2
the common difference(d1)
=a2-a1
=✓8 - ✓2
=2✓2-✓2
=✓2
the common difference(d2)
=a3-a2
=✓18 - ✓8
=3✓2-2✓2
=✓2
d1=d2
so the next three more term means that is
a5 =✓32+✓2=✓72
a6 =✓72+✓2=✓98
a7 =✓98+✓2=✓128
three more term is✓72,✓98,✓128 ans
(xiii) ✓3, ✓6, ✓9 , ✓12 , . . .
Check this squences are in AP
the first term(a)=✓3
the common difference(d1)
=a2-a1
=✓6 - ✓3
=✓3✓2-✓3
=✓3(✓2-1)
the common difference(d2)
=a3-a2
=✓9 - ✓6
=✓3✓3-✓3✓2
=✓3(✓3-✓2)
No, Because d1≠d2
(xiv) 1/ 2 , 3/2 , 5/2 , 7/2 , . . .
Check this squences are in AP
the first term(a)= 1/2
the common difference(d1)
=a2-a1
=3/2-1/2
=3/2 - 1/2 = 2/2
=1
the common difference(d2)
=a3-a2
=5/2-3/2
=2/2 =1
d1=d2
so the next three more term means that is
a5 =7/2+1=9/2
a6 =9/2+1=11/2
a7 =11/2+1=13/2
three more term is 9/2,11/2,13/2 ans
(xv) 1^2 , 5^2 , 7^2 , 73, . . .
1,25,49,73............
Check this squences are in AP
the first term(a)= 1
the common difference(d1)
=a2-a1
=25-1
=24
the common difference(d2)
=a3-a2
=49-25
=24
d1=d2
so the next three more term means that is
a5 =73+24=99
a6 =99+24=123
a7 =123+24=147
three more term is 99,123,147 ans
