Solution:

We will use the geometric progression method to solve this investment problem.

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Given:

• Initial investment $a = 200,000$
• Annual interest rate $r = 1 + \frac{5}{100} = 1.05$
• Number of years $n = 6$

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a) Total Amount After 6 Years:

The amounts at the end of each year form a geometric progression:

$A = a \cdot r^{n-1}$

The total amount $S$ after 6 years is the sum of this geometric progression:

$S = a + ar + ar^2 + \ldots + ar^{n-1}$

This is a geometric series where:

• First term $a = 200,000$
• Common ratio $r = 1.05$
• Number of terms $n = 6$

The sum of a geometric series is given by:

$S_n = \frac{a(r^n - 1)}{r - 1}$

Substitute the values:

$S_6 = \frac{200,000 \cdot (1.05^6 - 1)}{1.05 - 1}$

$S_6 = \frac{200,000 \cdot (1.3401 - 1)}{0.05}$

$S_6 = \frac{200,000 \cdot 0.3401}{0.05}$

$S_6 \approx 1,360,400$

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b) Explanation:

We used the geometric progression formula because the annual interest forms a sequence where each year's investment becomes the base for the next year's calculation, generating a multiplier effect. The yearly earnings compound, forming a geometric progression with a common ratio of 1.05.

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Would you like further breakdown or examples?

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Related Questions:

1. How is compound interest related to geometric progression?
2. What happens if the interest rate changes every year?
3. How can you use geometric progression for monthly compounding?
4. What if the investment was compounded semi-annually instead?
5. How do continuous compounding formulas differ from yearly compounding?

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Tip: Always double-check the interest rate and compounding period when applying the geometric progression formula.