Calculate the value of his investment at the end of the third year for each investment option, correct to two decimal places.

Determine Daniel’s best investment option.

Calculate u₁, the first term in the sequence.

Show that the common difference is 7.

S_n is the sum of the first n terms of the sequence.

Find an expression for S_n. Give your answer in the form S_n = An² + Bn, where A and B are constants.

The first term, v₁, of a geometric sequence is 20 and its fourth term v₄ is 67.5.

Show that the common ratio, r, of the geometric sequence is 1.5.

T_n is the sum of the first n terms of the geometric sequence.

Calculate T₇, the sum of the first seven terms of the geometric sequence.

T_n is the sum of the first n terms of the geometric sequence.

Use your graphic display calculator to find the smallest value of n for which T_n > S_n.

Option 1:     Amount    \( = 25\,000{\left( {1 + \frac{5}{{100}}} \right)^3}\)     (M1)(A1)

= \(28\,940.63\)     (A1)(G2)

Note: Award (M1) for substitution in compound interest formula, (A1) for correct substitution. Give full credit for use of lists.

Option 2:     Amount     \( = 25\,000{\left( {1 + \frac{{4.8}}{{12(100)}}} \right)^{3 \times 12}}\)     (M1)

= \(28\,863.81\)     (A1)(G2)

Note: Award (M1) for correct substitution in the compound interest formula. Give full credit for use of lists.

[8 marks]

Option 1 is the best investment option.     (A1)(ft)

[1 mark]

u₁ = 135 + 7(1)     (M1)

= 142     (A1)(G2)

[2 marks]

u₂ = 135 + 7(2) = 149     (M1)

d = 149 – 142     OR alternatives     (M1)(ft)

d = 7     (AG)

[2 marks]

\({S_n} = \frac{{n[2(142) + 7(n - 1)]}}{2}\)     (M1)(ft)

Note: Award (M1) for correct substitution in correct formula.

\( = \frac{{n[277 + 7n]}}{2}\)     OR equivalent     (A1)

\( = \frac{{7{n^2}}}{2} + \frac{{277n}}{2}\)     (= 3.5n² + 138.5n)     (A1)(G3)

[3 marks]

20r³ = 67.5     (M1)

r³ = 3.375     OR \(r = \sqrt[3]{{3.375}}\)     (A1)

r = 1.5     (AG)

[2 marks]

\({T_7} = \frac{{20({{1.5}^7} - 1)}}{{(1.5 - 1)}}\)     (M1)

Note: Award (M1) for correct substitution in correct formula.

= 643 (accept 643.4375)     (A1)(G2)

[2 marks]

\(\frac{{20({{1.5}^n} - 1)}}{{(1.5 - 1)}} > \frac{{7{n^2}}}{2} + \frac{{277n}}{2}\)     (M1)

Note: Award (M1) for an attempt using lists or for relevant graph.

n = 10     (A1)(ft)(G2)

Note: Follow through from their (c).

[2 marks]

For many, this question came as a welcome relief following the previous two questions. For those with a sound grasp of the topic, there were many very successful attempts.

A common error was to make all the comparisons using interest alone; though much credit was given for doing this, candidates should be aware of what is being asked for in the question.

Many did not understand the notion of monthly compounding periods.

For many, this question came as a welcome relief following the previous two questions. For those with a sound grasp of the topic, there were many very successful attempts.

A common error was to make all the comparisons using interest alone; though much credit was given for doing this, candidates should be aware of what is being asked for in the question.

Many did not understand the notion of monthly compounding periods.

For many, this question came as a welcome relief following the previous two questions. For those with a sound grasp of the topic, there were many very successful attempts.

A common weakness was seen in the “show that” parts of the question where, despite a lenient approach to method, many were unable to communicate their thoughts on paper.

For many, finding an expression for S_n in (c) was problematical.

The final part was challenging to the great majority, with a large number not attempting it at all; only the highly competent reached the correct answer.

For many, this question came as a welcome relief following the previous two questions. For those with a sound grasp of the topic, there were many very successful attempts.

A common weakness was seen in the “show that” parts of the question where, despite a lenient approach to method, many were unable to communicate their thoughts on paper.

For many, finding an expression for S_n in (c) was problematical.

The final part was challenging to the great majority, with a large number not attempting it at all; only the highly competent reached the correct answer.

For many, this question came as a welcome relief following the previous two questions. For those with a sound grasp of the topic, there were many very successful attempts.

A common weakness was seen in the “show that” parts of the question where, despite a lenient approach to method, many were unable to communicate their thoughts on paper.

For many, finding an expression for S_n in (c) was problematical.

The final part was challenging to the great majority, with a large number not attempting it at all; only the highly competent reached the correct answer.

For many, this question came as a welcome relief following the previous two questions. For those with a sound grasp of the topic, there were many very successful attempts.

A common weakness was seen in the “show that” parts of the question where, despite a lenient approach to method, many were unable to communicate their thoughts on paper.

For many, finding an expression for S_n in (c) was problematical.

The final part was challenging to the great majority, with a large number not attempting it at all; only the highly competent reached the correct answer.

For many, this question came as a welcome relief following the previous two questions. For those with a sound grasp of the topic, there were many very successful attempts.

A common weakness was seen in the “show that” parts of the question where, despite a lenient approach to method, many were unable to communicate their thoughts on paper.

For many, finding an expression for S_n in (c) was problematical.

The final part was challenging to the great majority, with a large number not attempting it at all; only the highly competent reached the correct answer.

For many, this question came as a welcome relief following the previous two questions. For those with a sound grasp of the topic, there were many very successful attempts.

A common weakness was seen in the “show that” parts of the question where, despite a lenient approach to method, many were unable to communicate their thoughts on paper.

For many, finding an expression for S_n in (c) was problematical.

The final part was challenging to the great majority, with a large number not attempting it at all; only the highly competent reached the correct answer.