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Question 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 Kilometre when the fare is ₹15 for the first km and ₹8 for each additional km.
(ii) The amount of air present in a cylinder when a vacuum pump removes ¼ of the air remaining in the cylinder at a time.
(iii) The cost of digging a well after every metre of digging, when it costs ₹150 for the first metre and rises by ₹50 for each subsequent metre.
(iv) The amount of money in the amount every year, when ₹10,000 is deposited at compound interest at 8% per annum.
Answer
(i)
First km (a1)=15
Second km (a2)=15+8=23
Third km(a3)=15+8×2=31
Fourth km (a4)=15+8×3=39
AP=15,23,31,39,……..
Common differences (d1)=23–15=8 , d2=31–23=8 , d3=39–31=8
In this AP common difference are same. Hence,it is form in AP.
(ii)
Let cylinder air (a1)=64 liter
Remaining air of 64 liter (a2)=64–¼×64=16 liter
Remaining air of 48 liter (a3)=48–48¼=36 liter
Remaining air of 36 liter (a4)=36–¼×36=25 liter
AP=64,48,36,25,………
d1=48–64 = –16 , d2=36–48 = –12 , d3=25–36 = –9
In this AP common difference are not same. Hence, it is not form in AP.
(iii)
First metre (a1)=₹150
Second metre (a2)=₹150+₹50=₹200
Third metre (a3)=₹150+₹50×2=₹250
Fourth metre (a4)=₹150+₹50×3=₹300
AP=150,200,250,300,……..
Common differences (d1)=200–150=50 , d2=250–200=50 , d3=300–250=50
In this AP common difference are same. Hence,it is form in AP.
(iv)
First year amount (a1)=₹10,000
Second year amount (a2)=₹10,000+8%of ₹10,000 = 10,000+8÷100×10,000=₹10,800
Third year amount (a3)=₹10,000+8%of ₹10,800 = 10,000+8÷100×10,800=₹11664
AP=10,000 , 10,800 , 11664 , …………..
Common differences (d1)=10,800–10,000=800 (d2)=11664–10,800=864
In this AP common difference are not same. Hence, it is not form in AP.
Question 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
Answer
a1=a
a1=10
a2=a+d
a2=10+10=20
a3=a+2d
a3=10+2×10=30
a4=a+3d
a4=10+3×10=40
AP=10,20,30,40
(ii) a= –2 , d=0
Answer
a1=a
a1= –2
a2=a+d
a2= –2+0= –2
a3=a+2d
a3= –2+2×0= –2
a4=a+3d
a4= –2+3×0= –2
AP= –2,–2,–2,–2
(iii) a= 4 , d= –3
Answer
a1=a
a1=4
a2=a+d
a2=4+(–3)=1
a3=a+2d
a3=4+2(–3)= –2
a4=a+3d
a4=4+3(–3)= –5
AP=4,1,–2,–5
(iv) a= –1 , d=1/2
Answer
a1=a
a1= –1
a2=a+d
a2= –1+1/2= –1/2
a3=a+2d
a3= –1+2×1/2= –1+1=0
a4=a+3d
a4= –1+3×1/2= –1+3/2=1/2
AP = –1,–1/2,0,1/2
(V) a= –1.25 , d= –0.25
Answer
a1=a
a1= –1.25
a2=a+d
a2= –1.25+(–0.25)= –1.50
a3=a+2d
a3= –1.25+2(–0.25)= –1.75
a4=a+3d
a4= –1.25+3(–0.25)= –2
AP= –1.25,–1.50,–1.75,–2
Question 3
For the following APs, write the first term and the common difference:
(i) 3,1,–1,–3, . . . . .
Answer
a1=a
a1=3
d=a2–a
d=1–3= –2
(ii) –5,–1,3,7, . . . . .
Answer
a1=a
a1= –5
d=a2–a
d1= –1–(–5)=4
(iii) 1/3 , 5/3 , 9/3 , 13/3 , . . . . . . .
Answer
a1=a
a1=1/3
d=a2–a
d=5/3–1/3=4/3
(iv) 0.6 , 1.7 , 2.8 , 3.9 , . . . . . . .
Answer
a1=a
a1=0.6
d=a2–a
d=1.7 –0.6=1.1
Question 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,…….
d1=a2–a
d1=4–2=2
d2=a3–a2
d2=8–4=4
d3=a4–a3
d3=16–8=8
No, it is not form in AP.
(ii) 2 , 5/2 , 3 , 7/2 , ……….
Answer
d1=a2–a
d1=5/2–2=1/2
d2=a3–a2
d2=3–5/2=1/2
d3=a4–a3
d3=7/2–3=1/2
Yes , it is form in AP.
a5=a+4d
a5=2+4×1/2=4
a6=a+5d
a6=2+5×1/2=9/2
a7=a+6d
a7=2+6×1/2=5
d=1/2;4,9/2,5
(iii) –1.2,–3.2,–5.2,–7.2,……..
Answer
d1=a2–a
d1= –3.2–(–1.2)= –2
d2=a3–a2
d2= –5.2–(–3.2)= –2
d3=a4–a3
d3= –7.2–(–5.2)= –2
Yes , it is form in AP.
a5=a+4d
a5= –1.2+4(–2)= –9.2
a6=a+5d
a6= –1.2+5(–2)= –11.2
a7=a+6d
a7= –1.2+6(–2)= –13.2
d= –2;–9.2,–11.2,–13.2
(iv) –10,–6,–2,2,….
Answer
d1=a2–a
d1= –6–(–10)=4
d2=a3–a2
d2= –2–(–6)=4
d3=a4–a3
d3=2–(–2)=4
Yes , it is form in AP.
a5=a+4d
a5= –10+4×4=6
a6=a+5d
a6= –10+5×4=10
a7=a+6d
a7= –10+6×4=14
d=4;6,10,14
(V) 3,3+√2,3+2√2,3+3√2,………..
Answer
d1=a2–a
d1=3+√2–3=√2
d2=a3–a2
d2=3+2√2–(3+√2)=3+2√2–3–√2=2√2–√2=√2(2–1)=√2
d3=a4–a3
d3=3+3√2–(3+2√2)=3+3√2–3–2√2=3√2–2√2
d3=√2(3–2)=√2
Yes , it is form in AP.
a5=a+4d
a5=3+4(√2)=3+4√2
a6=a+5d
a6=3+5(√2)=3+5√2
a7=a+6d
a7=3+6(√2)=3+6√2
d=√2;3+4√2,3+5√2,3+6√2
(vi) 0.2,0.22,0.222,0.2222,………
Answer
d1=a2–a
d1=0.22–0.2=0.02
d2=a3–a2
d2=0.222–0.22=0.002
No, it is not form in AP.
(vii) 0,–4,–8,–12,……..
Answer
d1=a2–a
d1= –4–0= –4
d2=a3–a2
d2= –8–(–4)= –4
d3=a4–a3
d3= –12–(–8)= –4
Yes , it is form in AP.
a5=a+4d
a5=0+4(–4)= –16
a6=a+5d
a6=0+5(–4)= –20
a7=a+6d
a7=0+6(–4)= –24
d= –4;–16,–20,–24
(viii) –1/2,–1/2,–1/2,–1/2,…….
Answer
d1=a2–a
d1= –1/2–(–1/2)=0
d2=a3–a2
d2= –1/2–(–1/2)=0
d3=a4–a3
d3= –1/2–(–1/2)=0
Common difference are zero. So, next terms are same.
d=0;–1/2,–1/2,–1/2
(ix) 1,3,9,27,……..
Answer
d1=a2–a
d1=3–1=2
d2=a3–a2
d2=9–3=6
No, it is not form in AP.
(x) a,2a,3a,4a,……
Answer
d1=2a–a=a
d2=3a–2a=a
d3=4a–3a=a
Yes , it is form in AP.
a5=a+4d
a5=a+4(a)=5a
a6=a+5d
a6=a+5(a)=6a
a7=a+6d
a7=a+6(a)=7a
d=a;5a,6a,7a
(xi) a,a²,a³,a⁴,…….
d1=a2–a
d1=a²–a=a(a–1)
d2=a3–a2
d2=a³–a²=a²(a–1)
No,it is not form in AP.
(xii) √2,√8,√18,√32,………
Answer
d1=a2–a
d1=√8–√2=√2(√4–1)=√2(2–1)=√2
d2=a3–a2
d2=√18–√8=√2(√9–√4)=√2(3–2)=√2
d3=a4–a3
d3=√32–√18=√2(√16–√9)=√2(4–3)=√2
Yes, it is form in AP.
a5=a+4d
a5=√2+4(√2)=√2+4√2=√2(1+4)=√2(5)
=√2×√25=√50
a6=a+5d
a6=√2+5(√2)=√2+5√2=√2(1+5)=√2(6)
=√2×√36=√72
a7=a+6d
a7=√2+6(√2)=√2+6√2=√2(1+6)=√2(7)
=√2×√49=√98
d=√2;√50,√72,√98
(xiii) √3,√6,√9,√12,…….
Answer
d1=a2–a
d1=√6–√3=√3(√2–1)
d2=a3–a2
d2=√9–√6=√3(√3–√2)
No,it is not form in AP.
(xiv) 1²,3²,5²,7²,……..
Answer
d1=a2–a
d1=3²–1²=9–1=8
d2=a3–a2
d2=5²–3²=25–9=16
No,it is not form in AP.
(xv) 1²,5²,7²,73,…….
Answer
d1=a2–a
d1=5²–1²=25–1=24
d2=a3–a2
d2=7²–5²=49–25=24
d3=a4–a3
d3=73–7²=73–49=24
Yes,it is form in AP.
a5=a+4d
a5=1²+4(24)=1+96=97
a6=a+5d
a6=1²+5(24)=1+120=121
a7=a+6d
a7=1²+6(24)=1+144=145
d=24;97,121,145
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