A sinking fund is set up with semi-annual payments of $1,500 at the end of each 6-month period into an account that earns 10% per annum compounded semi-annually. What is the minimum number of such payments required so that the accumulated amount is at least $21,000?

Introduction:
This question deals with the future value of an ordinary annuity. A person deposits 1,500 dollars every 6 months into an account that earns 10% annual interest, compounded semi-annually. We need to determine the smallest number of such payments that will accumulate to at least 21,000 dollars.

Given Data / Assumptions:

• Regular semi-annual payment, R = 1,500 dollars
• Nominal annual interest rate, 10% per annum
• Compounding frequency: semi-annually, so 2 periods per year
• Periodic rate per 6-month period, i = 10% / 2 = 5% = 0.05
• Target future value, S = 21,000 dollars
• Payments are made at the end of each 6-month period (ordinary annuity)

Concept / Approach:
The future value S of an ordinary annuity is given by: S = R * ((1 + i)^n - 1) / i where:

• R is the payment per period
• i is the interest rate per period
• n is the number of payments

We are given S, R, and i, and need the smallest integer n such that the formula yields S greater than or equal to 21,000 dollars.

Step-by-Step Solution:
Step 1: Note that i = 0.05 and R = 1500. Step 2: Write the expression for S as a function of n. S(n) = 1500 * ((1.05)^n - 1) / 0.05 Step 3: Check values of n starting from a reasonable small integer until S(n) ≥ 21000. For n = 10: S(10) = 1500 * ((1.05)^10 - 1) / 0.05 Compute (1.05)^10 ≈ 1.62889 S(10) ≈ 1500 * (1.62889 - 1) / 0.05 S(10) ≈ 1500 * 0.62889 / 0.05 ≈ 1500 * 12.5778 ≈ 18866.70 S(10) is less than 21000, so 10 payments are not enough. Step 4: Evaluate n = 11. (1.05)^11 ≈ 1.71033 S(11) = 1500 * (1.71033 - 1) / 0.05 S(11) ≈ 1500 * 0.71033 / 0.05 ≈ 1500 * 14.2066 ≈ 21309.90 S(11) is greater than 21000, so 11 payments are sufficient. Step 5: Since 10 is too small and 11 is enough, the minimum n is 11.

Verification / Alternative check:
Once we know that 11 payments are sufficient, it is useful to confirm that 10 payments do not reach the target. As shown, for 10 payments we only accumulate about 18,867 dollars. The jump to 11 payments increases both the amount of contributions and the interest, pushing the total above 21,000 dollars. This confirms that 11 is indeed the smallest integer that satisfies the requirement.

Why Other Options Are Wrong:
10 payments: The accumulated amount is about 18,867 dollars, which is less than 21,000 dollars. 12 and 13 payments: While these would also reach or exceed 21,000 dollars, they are not the minimum number of payments, as 11 already suffices. 14 payments: This would overshoot by an even larger margin and is clearly not minimal. 11 payments: This is the smallest n for which the accumulated value exceeds the target of 21,000 dollars.

Common Pitfalls:
Some learners attempt to solve the equation algebraically and end up with a non-integer value for n, then forget to round up to the next whole number. Others mistakenly treat the problem as simple interest or forget that payments occur at the end of each period. A further mistake is to misinterpret the rate by using 10% instead of 5% as the per period rate in the formula, which produces significantly incorrect results.

Final Answer:
The minimum number of semi-annual payments of 1,500 dollars required to accumulate at least 21,000 dollars at 10% per annum compounded semi-annually is 11 payments.