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maths..Arithmetic and Geometric Sequences (2)
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1)An earthworm starts moving towards a point 1.8cm away at a pace of 1cm per minute.As it gets tired,it covers only half the distance compared to previous minute in each succeeding minute.How long will it take to reach its target?(準確2 d.p.) as it gets tired, its speed is half the previous, so its speed is given by, (1cm per minute) *0.5^n = 0.5^n(cm per minute) distance travelled by the worm in successive minutes form a GP, =1cm, 0.5^1(cm per minute)*(1minute), 0.5^2(cm per minute)*(1minute),... , 0.5^(n-1)(cm per minute)*(1minute) =1cm, 0.5^1(cm), 0.5^2(cm),..., 0.5^(n-1)(cm) distance the worm need to travel: Sn=1+ 0.5^1+ 0.5^2+...+0.5^(n-1) =1.8 =>1*[1-0.5^n]/[1-0.5]=1.8 =>1-0.5^n=0.9 =>0.5^n=0.1 =>log 0.5^n= log 0.1 =>nlog 0.5= log 0.1 =>n= log 0.1/log 0.5 =>n=3.32 the worm need to go through the 4th minute and its speed at the 4th minute: 1*0.5^3 =0.125 cm/minute but after 3 minutes it has travelled (1+0.5+0.25)cm=1.75cm so the remaining distance for it: 1.8-1.75=0.05cm time taken for the final minute: 0.05cm/0.125 cm/minute =0.4 minute total time travelled: 3+0.4 =3.4 minutes 2)The sum of the first two terms of a geometric sequence is 54 and the sum to infinity is 72. Find the first term and the common ratio. plz write steps. let a be the first term and R be the common ratio the sum of GP is in the form, Sn=a + aR+ aR^2+... + aR^(n-1) put n=1, we get S1 =a + aR=54 a(1 + R)=54 a=54/(1 + R) ...(1) when n→∞, we get S∞=a(1 )/(1 - R)=72 a(1 )/(1 -R)=72 a/(1 - R)=72 ...(2) sub (1) into *(2) [54/(1 + R)]/(1 - R)=72 54/(1 - R^2)=72 3/(1 - R^2)=4 3=4(1 - R^2) 3=4 - 4R^2 R^2=1/4 R=1/2 or -1/2 put R into (1) when R=1/2 a=54/(1 + (1/2)) a=54/(3/2) a=36 when R=-1/2 a=54/(1 + (-1/2)) a=54/(1/2) a=108 in conclusion, (a=36 and R=1/2) or(a=108 and R=-1/2)
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maths..Arithmetic and Geometric Sequences (2)
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1)An earthworm starts moving towards a point 1.8cm away at a pace of 1cm per minute.As it gets tired,it covers only half the distance compared to previous minute in each succeeding minute.How long will it take to reach its target?(準確2 d.p.)2)The sum of the first two terms of a geometric sequence is 54 and the sum... 顯示更多 1)An earthworm starts moving towards a point 1.8cm away at a pace of 1cm per minute.As it gets tired,it covers only half the distance compared to previous minute in each succeeding minute.How long will it take to reach its target?(準確2 d.p.) 2)The sum of the first two terms of a geometric sequence is 54 and the sum to infinity is 72. Find the first term and the common ratio. plz write steps.最佳解答:
1)An earthworm starts moving towards a point 1.8cm away at a pace of 1cm per minute.As it gets tired,it covers only half the distance compared to previous minute in each succeeding minute.How long will it take to reach its target?(準確2 d.p.) as it gets tired, its speed is half the previous, so its speed is given by, (1cm per minute) *0.5^n = 0.5^n(cm per minute) distance travelled by the worm in successive minutes form a GP, =1cm, 0.5^1(cm per minute)*(1minute), 0.5^2(cm per minute)*(1minute),... , 0.5^(n-1)(cm per minute)*(1minute) =1cm, 0.5^1(cm), 0.5^2(cm),..., 0.5^(n-1)(cm) distance the worm need to travel: Sn=1+ 0.5^1+ 0.5^2+...+0.5^(n-1) =1.8 =>1*[1-0.5^n]/[1-0.5]=1.8 =>1-0.5^n=0.9 =>0.5^n=0.1 =>log 0.5^n= log 0.1 =>nlog 0.5= log 0.1 =>n= log 0.1/log 0.5 =>n=3.32 the worm need to go through the 4th minute and its speed at the 4th minute: 1*0.5^3 =0.125 cm/minute but after 3 minutes it has travelled (1+0.5+0.25)cm=1.75cm so the remaining distance for it: 1.8-1.75=0.05cm time taken for the final minute: 0.05cm/0.125 cm/minute =0.4 minute total time travelled: 3+0.4 =3.4 minutes 2)The sum of the first two terms of a geometric sequence is 54 and the sum to infinity is 72. Find the first term and the common ratio. plz write steps. let a be the first term and R be the common ratio the sum of GP is in the form, Sn=a + aR+ aR^2+... + aR^(n-1) put n=1, we get S1 =a + aR=54 a(1 + R)=54 a=54/(1 + R) ...(1) when n→∞, we get S∞=a(1 )/(1 - R)=72 a(1 )/(1 -R)=72 a/(1 - R)=72 ...(2) sub (1) into *(2) [54/(1 + R)]/(1 - R)=72 54/(1 - R^2)=72 3/(1 - R^2)=4 3=4(1 - R^2) 3=4 - 4R^2 R^2=1/4 R=1/2 or -1/2 put R into (1) when R=1/2 a=54/(1 + (1/2)) a=54/(3/2) a=36 when R=-1/2 a=54/(1 + (-1/2)) a=54/(1/2) a=108 in conclusion, (a=36 and R=1/2) or(a=108 and R=-1/2)
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