Geometric sequences, please help! 10 points

[From: ] [author: ] [Date: 12-02-18] [Hit: ]

01 = -(1/3)^n1/100 = (1/3)^nWe have to use logarithms on this partln(1/100) = n * ln(1/3)n = ln(1/100) / ln(1/3)n = -ln(100) / -ln(3)n = ln(100) / ln(3)n = 4.1918065485787692085931350440428So, at the 4th term, were just under 99% of 81.We want to go to the 5th term.Lets show it:54 + 54 * (1/3) + 54 * (1/3)^2 + 54 * (1/3)^3 + 54 * (1/3)^4 =>54 * (1 + (1/3) + (1/9) + (1/27) + (1/81)) =>54 * (81/81 + 27/81 + 9/81 + 3/81 + 1/81) =>(2/3) * (108 + 13) =>(2/3) * 121 =>242/3 =>80.......

S = 27 * 3
S = 81

Okay, you did well on the first 2 parts. The last part is cake:

S = ar * (r^n - 1) / (r - 1)
S = 0.99 * 81
0.99 * 81 = 54 * ((1/3)^n - 1) / (1/3 - 1)

All we have to do is solve for n

0.99 * 3 = 2 * ((1/3)^n - 1) / (-2/3)
0.99 * 3 = 2 * ((1/3)^n - 1) * (-3/2)
0.99 = ((1/3)^n - 1) * (-1)
0.99 = 1 - (1/3)^n
-0.01 = -(1/3)^n
1/100 = (1/3)^n

We have to use logarithms on this part

ln(1/100) = n * ln(1/3)
n = ln(1/100) / ln(1/3)
n = -ln(100) / -ln(3)
n = ln(100) / ln(3)
n = 4.1918065485787692085931350440428

So, at the 4th term, we're just under 99% of 81. We want to go to the 5th term. Let's show it:

54 + 54 * (1/3) + 54 * (1/3)^2 + 54 * (1/3)^3 + 54 * (1/3)^4 =>
54 * (1 + (1/3) + (1/9) + (1/27) + (1/81)) =>
54 * (81/81 + 27/81 + 9/81 + 3/81 + 1/81) =>
(2/3) * (108 + 13) =>
(2/3) * 121 =>
242/3 =>
80.666666

99% of 81 is 80.19

There you go.

-

an=a1*r^(n-1)

a4=a1*r^3

2=54*r^3

r^3=2/54=1/27

r=(1/27)^1/3=1/3

Sn=a1/(1-r)

=54/(1- 1/3)=54/(2/3)=81 correct on both

Sn>0.99Sinf

Sn>0.99(81)=80.19

a1(1-r^n)/(1-r) >80.19

54(1-(1/3)^n)/2/3 > 80.19

54 - 54(1/3)^n >53.46

-54(1/3)^n >-0.54

(1/3)^n >0.01

n>ln(0.01)/ln(1/3)

n>4.2 approx

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