Geometric Series: Salary with Annual Rises

A quantity multiplied by the same factor each step is a geometric progression. A salary rising $5\%$ a year is multiplied by $1.05$ annually, so $r = 1.05$. Use $ar^{n-1}$ for one year and $\frac{a(r^n - 1)}{r - 1}$ for a total.

A graduate’s salary is $\text{RM}30{,}000$ in the first year and increases by $5\%$ each year. Find (a) the salary in the tenth year, and (b) the total earned over the first ten years. Give answers to the nearest RM. [6]

The working

With $a = 30{,}000$ and $r = 1.05$:

Part (a), tenth-year salary is the $10$th term, $ar^{n-1}$ with $n = 10$: $T_{10} = 30{,}000\,(1.05)^{9} = 30{,}000\,(1.5513) \approx \text{RM}46{,}540 \quad \text{(M1, A1)}$

Part (b), total over ten years is the sum $S_{10} = \dfrac{a(r^{n} - 1)}{r - 1}$ with $n = 10$: $S_{10} = \frac{30{,}000\,\big((1.05)^{10} - 1\big)}{1.05 - 1} \quad \text{(M1)}$ $= \frac{30{,}000\,(1.6289 - 1)}{0.05} = \frac{30{,}000\,(0.6289)}{0.05} \quad \text{(M1)}$ $\approx \text{RM}377{,}337 \quad \text{(A1, A1)}$

Where the marks are won and lost

• The ratio is $r = 1.05$, not $0.05$: a $5\%$ rise multiplies by $1.05$.
• For the tenth term, the power is $n - 1 = 9$, not $10$: the first year uses $r^0$.
• For the total, use the sum formula, not the term formula. Reading “in the tenth year” versus “over ten years” decides which one.

Common mistakes

• Treating it as arithmetic (adding RM1500 a year) instead of geometric.
• Using $r^{10}$ for the tenth term (off-by-one in the power).
• Mixing up the term and sum formulas.

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