? asked in 科學及數學數學 · 1 decade ago

Arithmetic sequence$$#

Since 1 January 2000 , Mr.Wong has been investing 5% of his monthly salaries in a fund each month for the eduation of his daughter . Suppose his monthly salary was $40000 in 2000 and his monthly salary increases by 4% each year , in which year will the money invested be greater than $ 36000 ?

Update:

唔係A.P .......係 G.P ///

2 Answers

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  • 1 decade ago
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    In 2000 (first year), the invesment made by Mr. Wong is given by:

    $40000 × 5% = $2000

    So using the common ratio 1.04 (since his monthly salary increases by 4% each year), we can see that"

    Investment in 2001 (2nd year) = 2000 × 1.04

    Investment in 2002 (3rd year) = 2000 × 1.042

    Investment in (n th year) = 2000 × 1.04n

    Then at the nth year counting from 2000, the total investment made is given by:

    S(n) = a(Rn - 1)/(R - 1)

    = 2000(1.04n - 1)/0.04

    So for the total investment to be greater than $36000, we have:

    2000(1.04n - 1)/0.04 > 36000

    1.04n - 1 > 0.72

    1.04n > 1.72

    n × log (1.04) > log (1.72)

    n > log (1.72)/log (1.04)

    = 13.82

    So, n = 14 since n must be an integer.

    Therefore, at the year of 2013 (counting for the 14th year from 2000), the investment will be more than $36000.

    Source(s): My Maths knowledge
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  • ?
    Lv 7
    1 decade ago

    Since 1 January 2000 , Mr.Wong has been investing 5% of his monthly salaries in a fund each month for the eduation of his daughter . Suppose his monthly salary was $40000 in 2000 and his monthly salary increases by 4% each year , in which year will the money invested be greater than $ 36000 ?

    黃先生在第一年的投資為 40000x5% = $2000

    第 n 年,每月的投資為

    2000(1 + 4%)n-1

    依題意

    2000(1 + 4%)n-1 > 36000

    (1.04)n-1 > 18

    log(1.04)n-1 > log18

    n – 1 > log18 / log1.04

    n – 1 > 73.7

    n > 74.7

    所以在75年後(2075年),它的投資會大於$36000

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