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Date May 2010 Marks available 2 Reference code 10M.2.sl.TZ1.5
Level SL only Paper 2 Time zone TZ1
Command term Calculate Question number 5 Adapted from N/A

Question

Daniel wants to invest \(\$ 25\,000\) for a total of three years. There are two investment options.

Option One     pays compound interest at a nominal annual rate of interest of 5 %, compounded annually.

Option Two     pays compound interest at a nominal annual rate of interest of 4.8 %, compounded monthly.

An arithmetic sequence is defined as

un = 135 + 7n,     n = 1, 2, 3, …

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

[8]
A.a.

Determine Daniel’s best investment option.

[1]
A.b.

Calculate u1, the first term in the sequence.

[2]
B.a.

Show that the common difference is 7.

[2]
B.b.

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

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

[3]
B.c.

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.

[2]
B.d.

Tn 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.

[2]
B.e.

Tn 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 Tn > Sn.

[2]
B.f.

Markscheme

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]

A.a.

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


[1 mark]

A.b.

u1 = 135 + 7(1)     (M1)

= 142     (A1)(G2)

[2 marks]

B.a.

u2 = 135 + 7(2) = 149     (M1)

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

d = 7     (AG)

[2 marks]

B.b.

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

B.c.

20r= 67.5     (M1)

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

r = 1.5     (AG)

[2 marks]

B.d.

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


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


= 643 (accept 643.4375)     (A1)(G2)

[2 marks]

B.e.

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

B.f.

Examiners report

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.

A.a.

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.

A.b.

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 Sn 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.

B.a.

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 Sn 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.

B.b.

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 Sn 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.

B.c.

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 Sn 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.

B.d.

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 Sn 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.

B.e.

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 Sn 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.

B.f.

Syllabus sections

Topic 1 - Number and algebra » 1.8 » Use of the formulae for the \(n\)th term and the sum of the first \(n\) terms of the sequence.
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