WJEC Unit 3 Specimen — Question 5 5 marks

Exam BoardWJEC
ModuleUnit 3 (Unit 3)
SessionSpecimen
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeometric Sequences and Series
TypeTotal over time period
DifficultyStandard +0.3 This is a straightforward geometric series application with clear formula given. Students must recognize the pattern extends to 30 terms, apply the GP sum formula S_n = a(r^n - 1)/(r - 1), and perform calculator arithmetic. Slightly above average due to the 5-mark allocation and need to handle large powers, but requires no novel insight—it's a standard textbook exercise on geometric series in a financial context.
Spec1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1

Aled decides to invest £1000 in a savings scheme on the first day of each year. The scheme pays 8% compound interest per annum, and interest is added on the last day of each year. The amount of savings, in pounds, at the end of the third year is given by $$1000 \times 1 \cdot 08 + 1000 \times 1 \cdot 08^2 + 1000 \times 1 \cdot 08^3$$ Calculate, to the nearest pound, the amount of savings at the end of thirty years. [5]

AnswerMarks Guidance
After 30 years, saving is \((1.08)1000 + (1.08)^2 1000 + ... + (1.08)^{30}1000\)B1
This is G.P with \(a = (1.08)1000\)B2 (B2 for 3 correct, B1 for 2 correct)
\(r = 1.08\) and \(n = 30\)
Then \(S_{30} = (1000)(1.08)\left(\frac{(1.08)^{30}-1}{0.08}\right)\)M1 (correct formula)
\(\approx £122,346\)A1
Total: [5]
After 30 years, saving is $(1.08)1000 + (1.08)^2 1000 + ... + (1.08)^{30}1000$ | B1 |

This is G.P with $a = (1.08)1000$ | B2 | (B2 for 3 correct, B1 for 2 correct)

$r = 1.08$ and $n = 30$ | |

Then $S_{30} = (1000)(1.08)\left(\frac{(1.08)^{30}-1}{0.08}\right)$ | M1 | (correct formula)

$\approx £122,346$ | A1 |

**Total: [5]**
Aled decides to invest £1000 in a savings scheme on the first day of each year. The scheme pays 8% compound interest per annum, and interest is added on the last day of each year. The amount of savings, in pounds, at the end of the third year is given by
$$1000 \times 1 \cdot 08 + 1000 \times 1 \cdot 08^2 + 1000 \times 1 \cdot 08^3$$

Calculate, to the nearest pound, the amount of savings at the end of thirty years. [5]

\hfill \mbox{\textit{WJEC Unit 3  Q5 [5]}}