English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

please solve these questions for me.. thanks


geometric progression

calculate the number of terms of the geometric progression 1/4,1/12, 1/36....,1/324.


for the geometric progression 500,100,20... which term will be less than 1/10 for the first time?


how many terms of the geometric progression 48,24,12,...must be equal for the sum to be equal to 95/1/4?


the third term and the 6th term of a geometric progression are 27 and 8. find the second term.

the amount if the insurance coverage for a newly bought imported car in the year 2001 was rm120 000. the amount insured for each of the following years will decrease by 5% as compared to d previous year. calculate the amount insured for the year 2009.

-------------------------------

sum of infinity of a geometric progression

the second term n the 4th term of a geometric prgression are 20 and 12/4/5 respectively. if all the terms are positive, find the sum to infinity.

2007-02-04 01:29:48 · 2 answers · asked by Anonymous in Science & Mathematics Mathematics

2 answers

1) 1/4, 1/12, 1/36, .... , 1/324

To calculate the number of terms, we just need to know what term
1/324 is.

In this geometric progression, a1 = 1/4, and r = (1/12) / (1/4) = 1/3.

Therefore,
a[n] = (a1)r^(n - 1)

We just need to solve for n, letting a[n] = 1/324, a1 = 1/4, and
r = 1/3.

1/324 = (1/4)(1/3)^(n - 1)

Multiplying both sides by 4,

4/324 = (1/3)^(n - 1)

Reducing the fraction,

1/81 = (1/3)^(n - 1)

We can actually solve this without using logarithms, since 1/81 is equal to (1/3)^4.

(1/3)^4 = (1/3)^(n - 1)

Now, equate the exponents.

4 = n - 1, which gives us
n = 5. There are 5 terms in the geometric progression.

2) 500, 100, 20, .....

For this series, a1 = 500, and r = 100/500 = 1/5. That makes the formula
a[n] = (a1)r^(n - 1)
a[n] = (500)(1/5)^(n - 1)

We want to solve for a[n] such that a[n] < 1/10, so we solve the inequality

a[n] < 1/10
(500)(1/5)^(n - 1) < 1/10
Dividing both sides by 500, we get

(1/5)^(n - 1) < 1/5000

Take the antilog of both sides,

n - 1 < log[base 1/5] (1/5000)

Solve this inequality for n, take the nearest integer.

2007-02-04 01:40:02 · answer #1 · answered by Puggy 7 · 0 0

I am not really good at this but i give it a try let's see. I know 4 sure d first one is 1/108 (cuz u multiply everyone by 3) I think d next one is 5 (cuz u divide each one by 5) don't kno bout d rest sorry but i made myself feel good cuz i never did dis in school but i shocked myself by answering d first 2 questions BUT d first guy cud b right it seems like he knows what he's tlkin about goo luck

2007-02-04 09:40:51 · answer #2 · answered by Anonymous · 0 0

fedest.com, questions and answers