Derive and use the future value formula for an annuity to find the accumulated value of a series of equal regular contributions

Per-period rate: $r = \frac{0.06}{12} = 0.005$. Payment: $M = 200$. Number of payments: $n = 60$.

The future value of an ordinary annuity is

$A = M \cdot \frac{(1 + r)^n - 1}{r} = 200 \cdot \frac{(1.005)^{60} - 1}{0.005}$.

$(1.005)^{60} \approx 1.34885$, so $(1.005)^{60} - 1 \approx 0.34885$.

$A \approx 200 \cdot \frac{0.34885}{0.005} = 200 \cdot 69.770 \approx 13954.01$, i.e. $13954.01.

Markers reward the per-period rate, the right formula, an accurate compound factor, and a final answer to cents.

$r = \frac{0.048}{12} = 0.004$, $n = 120$, $A = 50000$.

Rearrange the future-value formula for $M$:

$M = \frac{A \cdot r}{(1 + r)^n - 1} = \frac{50000 \cdot 0.004}{(1.004)^{120} - 1}$.

$(1.004)^{120} \approx 1.614528$, so the denominator is $0.614528$.

$M = \frac{200}{0.614528} \approx 325.45$, i.e. $325.45.

Markers expect the correct per-period rate, the rearrangement for $M$, an accurate compound factor, and a final answer to cents.

Set up the per-period quantities. The annual rate is compounded monthly, so

Apply the future-value formula.

With $(1.005)^{24} \approx 1.12716$, the numerator is $0.12716$, so

i.e. $7629.59.

Marker's note: one mark for the per-period rate $r = 0.005$ with $n = 24$, one for substituting into $A = M\frac{(1+r)^n - 1}{r}$ and reaching $7629.59 to the nearest cent. Using the annual rate $0.06$ or $n = 2$ (years) loses the method mark.

Identify the inputs. Payments are annual and the rate is compounded annually, so

Compute the future value.

Since $(1.07)^6 \approx 1.50073$, the numerator is $0.50073$, giving

i.e. $10729.94.

Interest earned is future value minus total deposited. The total put in is $Mn = 1500 \times 6 = 9000$, so

i.e. $1729.94.

Marker's note: one mark for the future value $10729.94, one for computing total deposits $Mn = 9000$, one for the interest $1729.94. Quoting $A$ itself as the interest is the classic slip.

Set up and rearrange for $n$. Here $r = \frac{0.06}{12} = 0.005$, $M = 400$ and the target is $A = 20000$. Rearranging the future-value formula,

Substitute. First the bracket:

Then

Round up to a whole deposit. The balance passes $20000 partway through the 45th month, so $45$ deposits are needed.

Marker's note: one mark for the rearrangement and the bracket value $1.25$, one for $n \approx 44.74$ from the logarithms, one for rounding up to $45$. Rounding to $44$ (or truncating) leaves the balance short of $20000 and loses the final mark.

Part (a): read the timing from the ordinary-annuity model. Each deposit is made at the end of its year. Deposit 1 lands at the end of year $1$, so between then and the end of year $3$ only $3 - 1 = 2$ full years pass. That is why it compounds for $2$ years, and why the final deposit (end of year $3$) earns nothing.

Part (b): add the grown values in the table.

i.e. $8116.00.

Confirm with the formula using $r = 0.08$, $M = 2500$, $n = 3$:

i.e. $8116.00. The table sum and the formula agree.

Marker's note: one mark for the end-of-year timing explanation in (a), one for summing the table to $8116.00, one for a correct formula substitution, one for the matching $8116.00. A candidate who says deposit 1 compounds for $3$ years has misread the ordinary-annuity convention and loses part (a).

Set up and rearrange for the payment. The target and inputs are

and the required payment is

Substitute. With $(1.004)^{180} \approx 2.05148$, the denominator is $1.05148$, so

i.e. each monthly deposit must be about $304.33.

Marker's note: one mark for $r = 0.004$ and $n = 180$, one for the rearrangement $M = \frac{Ar}{(1+r)^n - 1}$, one for $304.33 to the nearest cent. Dividing by $(1+r)^n$ instead of $(1+r)^n - 1$ is a single-lump-sum discount and is wrong here.

Part (a): each payment earns one extra period of interest. In an annuity due every deposit is made at the start of its period rather than the end, so each one sits in the account one full period longer than in the ordinary model. Multiplying every term of the ordinary-annuity sum by $(1 + r)$ accounts for that single extra period, so

Part (b): compute the ordinary value, then adjust. Here

With $(1.005)^{480} \approx 10.95745$, the numerator is $9.95745$, so the ordinary value is

Multiply by $(1 + r)$ for the annuity due:

i.e. about $500362.

Marker's note: one mark for the "one extra period" reasoning in (a), one for $r = 0.005$ and $n = 480$, one for the ordinary value $497872.68, one for multiplying by $1.005$ to reach $500362. Forgetting the $(1 + r)$ factor caps the answer at the ordinary value.

Part (a): the deposit phase is an ordinary annuity. Quarterly compounding gives

Apply the future-value formula:

With $(1.02)^{20} \approx 1.48595$, the numerator is $0.48595$, so

which is about $14578, as required.

Part (b): no more deposits, so the balance grows as a single lump sum. For the next $3$ years there are $3 \times 4 = 12$ more quarterly compoundings and no annuity payments, so

With $(1.02)^{12} \approx 1.26824$,

i.e. about $18489.

Marker's note: one mark for the quarterly inputs $r = 0.02$, $n = 20$, one for the annuity sum reaching $14578 in (a); one mark for recognising the second phase as lump-sum compound growth (not another annuity), one for the factor $(1.02)^{12}$, one for the final $18489. Treating the second phase as a continuing annuity, or compounding for the wrong number of quarters, is the trap.