Unit 2 Review Exercises

Sequences and Series

These questions mix the ideas from the whole unit. Paper 1 means no calculator; Paper 2 means a GDC is allowed.

1 Paper 1 The 3rd term of an arithmetic sequence is \(11\) and the 8th term is \(31\). Find the first term, the common difference, and the sum of the first 20 terms.

\(u_1 + 2d = 11\), \(u_1 + 7d = 31 \Rightarrow 5d = 20\), so \(d = 4\) and \(u_1 = 3\).

\(S_{20} = \dfrac{20}{2}(6 + 19 \times 4) = 10 \times 82 = 820\). See 2.2.

2 Paper 1 A geometric sequence has first term \(48\) and common ratio \(-\dfrac{1}{2}\). Find the 6th term and the sum to infinity.

\(u_6 = 48 \times \left(-\dfrac{1}{2}\right)^5 = -\dfrac{48}{32} = -\dfrac{3}{2}\).

\(S_\infty = \dfrac{48}{1 + \frac{1}{2}} = 32\). See 2.3 and 2.4.

3 Paper 1 Find \(\displaystyle\sum_{r=1}^{30} (3r + 1)\).

Arithmetic, first term \(4\), last term \(91\), 30 terms: \(\;15 \times 95 = 1425\). See 2.6.

4 Paper 1 The numbers \(k\), \(k + 6\) and \(4k\) are consecutive terms of a geometric sequence. Find the possible values of \(k\), and the common ratio in each case.

\(\dfrac{k + 6}{k} = \dfrac{4k}{k + 6} \Rightarrow (k + 6)^2 = 4k^2 \Rightarrow k + 6 = 2k\) or \(k + 6 = -2k\).

\(k = 6\): terms \(6, 12, 24\), ratio \(2\). \(\qquad k = -2\): terms \(-2, 4, -8\), ratio \(-2\). See 2.3.

5 Paper 1 The sum of the first \(n\) terms of a series is \(S_n = 3n^2 - n\). Find \(u_n\), and show that the series is arithmetic.

\(u_n = S_n - S_{n-1} = 3n^2 - n - \big(3(n - 1)^2 - (n - 1)\big) = 6n - 4\) (and \(u_1 = S_1 = 2\) ✓).

\(u_{n+1} - u_n = 6\), a constant, so the series is arithmetic. See 2.1 and 2.2.

6 Paper 2 $6000 is invested at 3.75% per year, compounded monthly.

(a) Find the value of the investment after 7 years.

(b) Find the number of complete years needed for the value to reach $9000.

(a) \(6000\left(1 + \frac{3.75}{1200}\right)^{84} = \$7797.87\).

(b) TVM solver for N (PV \(= -6000\), FV \(= 9000\), I \(= 3.75\), PpY = CpY \(= 12\)): N \(= 129.95\ldots\) months, that is \(10.8\) years. So 11 complete years. See 2.5.

7 Paper 2 A company offers two salary plans, each starting at $40 000 in the first year. Plan A increases the salary by $1500 each year. Plan B increases it by 3.5% each year.

(a) Find the total amount earned in the first 10 years under each plan.

(b) In which year is the Plan B salary first higher than the Plan A salary?

(a) Plan A (arithmetic): \(\dfrac{10}{2}(80\,000 + 9 \times 1500) = \$467\,500\).

Plan B (geometric): \(\dfrac{40\,000\left(1.035^{10} - 1\right)}{0.035} = \$469\,256\) (nearest dollar).

(b) Compare \(40\,000 \times 1.035^{n-1}\) with \(40\,000 + 1500(n - 1)\) in a GDC table. Year 5: B \(= 45\,901\), A \(= 46\,000\). Year 6: B \(= 47\,507\), A \(= 47\,500\). So in the 6th year. See 2.2 and 2.3.

8 Paper 1 Consider the geometric series \(1 + (2x - 1) + (2x - 1)^2 + \cdots\)

(a) Find the values of \(x\) for which the series converges.

(b) Find \(x\) when the sum to infinity is \(3\).

(a) \(|2x - 1| < 1 \Rightarrow -1 < 2x - 1 < 1 \Rightarrow 0 < x < 1\).

(b) \(\dfrac{1}{1 - (2x - 1)} = \dfrac{1}{2 - 2x} = 3 \Rightarrow 2 - 2x = \dfrac{1}{3} \Rightarrow x = \dfrac{5}{6}\), inside the range ✓. See 2.4.

9 Paper 1 Write \(0.1\dot{8} = 0.1888\ldots\) as a fraction in its lowest terms.

\(0.1 + (0.08 + 0.008 + \cdots) = \dfrac{1}{10} + \dfrac{0.08}{1 - 0.1} = \dfrac{1}{10} + \dfrac{8}{90} = \dfrac{17}{90}\). See 2.4.

10 Paper 2 A lake contains 200 tonnes of a pollutant. Each year 10% of it is removed, and then 30 tonnes are added. Let \(u_n\) be the amount at the start of year \(n\), with \(u_1 = 200\).

(a) Write down a recurrence relation and find \(u_4\).

(b) Use your GDC to describe what happens to \(u_n\) in the long term.

(a) \(u_{n+1} = 0.9\,u_n + 30\). \(\;u_2 = 210\), \(u_3 = 219\), \(u_4 = 227.1\) tonnes.

(b) A GDC table shows \(u_n\) increasing more and more slowly towards \(300\) tonnes. (Check: if the amount stays at \(L\), then \(L = 0.9L + 30\), so \(L = 300\).) See 2.1.