# 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 ?

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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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• 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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