IB DP Mathematical Studies 1.7 Use of the formulae for the nth term and the sum of the first n terms of the sequence Paper 2

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Question

The natural numbers: 1, 2, 3, 4, 5… form an arithmetic sequence.

A geometric progression $G_1$ has 1 as its first term and 3 as its common ratio.

State the values of u₁ and d for this sequence.[2]

i.a.

Use an appropriate formula to show that the sum of the natural numbers from 1 to n is given by $\frac{1}{2}n (n +1)$.[2]

i.b.

Calculate the sum of the natural numbers from 1 to 200.[2]

i.c.

The sum of the first n terms of G₁ is 29 524. Find n.[3]

ii.a.

A second geometric progression G₂ has the form $1,\frac{1}{3},\frac{1}{9},\frac{1}{{27}}…$[1]

ii.b.

Calculate the sum of the first 10 terms of G₂.[2]

ii.c.

Explain why the sum of the first 1000 terms of G₂ will give the same answer as the sum of the first 10 terms, when corrected to three significant figures.[1]

ii.d.

Using your results from parts (a) to (c), or otherwise, calculate the sum of the first 10 terms of the sequence $2,3\frac{1}{3},9\frac{1}{9},27\frac{1}{{27}}…$

Give your answer correct to one decimal place.[3]

ii.e.

Answer/Explanation

Markscheme

$u_1 = d = 1$. (A1)(A1)[2 marks]

i.a.

Sum is $\frac{1}{2}n(2{u_1} + d(n – 1))$ or $\frac{1}{2}n({u_1} + {u_n})$ (M1)

Award (M1) for either sum formula seen, even without substitution.

So sum is $\frac{1}{2}n(2 + (n – 1)) = \frac{1}{2}n(n + 1)$ (A1)(AG)

Award (A1) for substitution of ${u_1} = 1 = d$ or ${u_1} = 1$ and ${u_n} = n$ with simplification where appropriate. $\frac{1}{2}n(n + 1)$ must be seen to award this (A1).[2 marks]

i.b.

$\frac{1}{2}(200)(201) = 20 100$ (M1)(A1)(G2)

(M1) is for correct formula with correct numerical input. Original sum formula with u, d and n can be used.[2 marks]

i.c.

$\frac{{1 – {3^n}}}{{1 – 3}} = 29524$ (M1)(A1)

(M1) for correctly substituted formula on one side, (A1) for = 29524 on the other side.

n = 10. (A1)(G2)

Trial and error is a valid method. Award (M1) for at least $\frac{{1 – {3^{10}}}}{{1 – 3}}$ seen and then (A1) for = 29524, (A1) for $n = 10$. For only unproductive trials with $n \ne 10$, award (M1) and then (A1) if the evaluation is correct.[3 marks]

ii.a.

Common ratio is $\frac{1}{3}$, (0.333 (3sf) or 0.3) (A1)

Accept ‘divide by 3’.[1 mark]

ii.b.

$\frac{{1 – {{\left( {\frac{1}{3}} \right)}^{10}}}}{{1 – \frac{1}{3}}}$ (M1)

= 1.50 (3sf) (A1)(ft)(G1)

1.5 and $\frac{3}{2}$ receive (A0)(AP) if AP not yet used Incorrect formula seen in (a) or incorrect value in (b) can follow through to (c). Can award (M1) for $1 + \left( {\frac{1}{3}} \right) + \left( {\frac{1}{9}} \right) + ……$[2 marks]

ii.c.

Both ${\left( {\frac{1}{3}} \right)^{10}}$ and ${\left( {\frac{1}{3}} \right)^{1000}}$ (or those numbers divided by 2/3) are 0 when corrected to 3sf, so they make no difference to the final answer. (R1)

Accept any valid explanation but please note: statements which only convey the idea of convergence are not enough for (R1). The reason must show recognition that the convergence is adequately fast (though this might be expressed in a much less technical manner).[1 mark]

ii.d.

The sequence given is $G_1 + G_2$ (M1)

The sum is 29 524 + 1.50 (A1)(ft)

= 29 525.5 (A1)(ft)(G2)

The (M1) is implied if the sum of the two numbers is seen. Award (G1) for 29 500 with no working. (M1) can be awarded for
$2 + 3\frac{1}{3} + …$ Award final (A1) only for answer given correct to 1dp.[3 marks]

ii.e.

Question

Give all answers in this question correct to the nearest dollar.

Clara wants to buy some land. She can choose between two different payment options. Both options require her to pay for the land in 20 monthly installments.

Option 1: The first installment is $\$ 2500$. Each installment is $\$ 200$ more than the one before.

Option 2: The first installment is $\$ 2000$. Each installment is $8\% $ more than the one before.

If Clara chooses option 1,

(i) write down the values of the second and third installments;

(ii) calculate the value of the final installment;

(iii) show that the total amount that Clara would pay for the land is $\$ 88000$.[7]

If Clara chooses option 2,

(i) find the value of the second installment;

(ii) show that the value of the fifth installment is $\$ 2721$.[4]

The price of the land is $\$ 80000$. In option 1 her total repayments are $\$ 88000$ over the 20 months. Find the annual rate of simple interest which gives this total.[4]

Clara knows that the total amount she would pay for the land is not the same for both options. She wants to spend the least amount of money. Find how much she will save by choosing the cheaper option.[4]

Answer/Explanation

Markscheme

(i) Second installment $ = \$ 2700$ (A1)

Third installment $ = \$ 2900$ (A1)

(ii) Final installment $ = 2500 + 200 \times 19$ (M1)(A1)

Note: (M1) for substituting in correct formula or listing, (A1) for correct substitutions.

$ = \$ 6300$ (A1)(G2)

(iii) Total amount $ = \frac{{20}}{2}(2500 + 6300)$

$ \frac{{20}}{2}(5000 + 19 \times 200)$ (M1)(A1)

Note: (M1) for substituting in correct formula or listing, (A1) for correct substitution.

$ = \$ 88000$ (AG)

Note: Final line must be seen or previous (A1) mark is lost.[7 marks]

(i) Second installment $2000 \times 1.08 = \$ 2160$ (M1)(A1)(G2)

Note: (M1) for multiplying by $1.08$ or equivalent, (A1) for correct answer.

(ii) Fifth installment $ = 2000 \times {1.08^4} = 2720.98 = \$ 2721$ (M1)(A1)(AG)

Notes: (M1) for correct formula used with numbers from the problem. (A1) for correct substitution. The $2720.9 \ldots $ must be seen for the (A1) mark to be awarded. Accept list of 5 correct values. If values are rounded prematurely award (M1)(A0)(AG).[4 marks]

Interest is $ = \$ 8000$ (A1)

$80000 \times \frac{r}{{100}} \times \frac{{20}}{{12}} = 8000$ (M1)(A1)

Note: (M1) for attempting to substitute in simple interest formula, (A1) for correct substitution.

Simple Interest Rate $ = 6\% $ (A1)(G3)

Note: Award (G3) for answer of $6\% $ with no working present if interest is also seen award (A1) for interest and (G2) for correct answer.[4 marks]

Financial accuracy penalty (FP) is applicable where indicated in the left hand column.

(FP) Total amount for option 2 $ = 2000\frac{{(1 – {{1.08}^{20}})}}{{(1 – 1.08)}}$ (M1)(A1)

Note: (M1) for substituting in correct formula, (A1) for correct substitution.

$ = \$ 91523.93$ ($ = \$ 91524$) (A1)
$91523.93 – 88000 = \$ 3523.93 = \$ 3524$ to the nearest dollar (A1)(ft)(G3)

Note: Award (G3) for an answer of $\$ 3524$ with no working. The difference follows through from the sum, if reasonable. Award a maximum of (M1)(A0)(A0)(A1)(ft) if candidate has treated option 2 as an arithmetic sequence and has followed through into their common difference. Award a maximum of (M1)(A1)(A0)(ft)(A0) if candidate has consistently used $0.08$ in (b) and (d).[4 marks]

Question

Throughout this question all the numerical answers must be given correct to the nearest whole number.

Park School started in January 2000 with $100$ students. Every full year, there is an increase of $6\% $ in the number of students.

Find the number of students attending Park School in

(i) January 2001;

(ii) January 2003.[4]

Park School started in January 2000 with $100$ students. Every full year, there is an increase of $6\% $ in the number of students.

Show that the number of students attending Park School in January 2007 is $150$.[2]

Grove School had $110$ students in January 2000. Every full year, the number of students is $10$ more than in the previous year.

Find the number of students attending Grove School in January 2003.[2]

Grove School had $110$ students in January 2000. Every full year, the number of students is $10$ more than in the previous year.

Find the year in which the number of students attending Grove School will be first $60\% $ more than in January 2000.[4]

Each January, one of these two schools, the one that has more students, is given extra money to spend on sports equipment.

(i) Decide which school gets the money in 2007. Justify your answer.

(ii) Find the first year in which Park School will be given this extra money.[5]

Answer/Explanation

Markscheme

(i) $100 \times 1.06 = 106$ (M1)(A1)(G2)

Note: (M1) for multiplying by $1.06$ or equivalent. (A1) for correct answer.

(ii) $100 \times {1.06^3} = 119$ (M1)(A1)(G2)

Note: (M1) for multiplying by ${1.06^3}$ or equivalent or for list of values. (A1) for correct answer.[4 marks]

$100 \times {1.06^7} = 150.36 \ldots = 150$ correct to the nearest whole (M1)(A1)(AG)

Note: (M1) for correct formula or for list of values. (A1) for correct substitution or for $150$ in the correct position in the list. Unrounded answer must be seen for the (A1).[2 marks]

$110 + 3 \times 10 = 140$ (M1)(A1)(G2)

Note: (M1) for adding $30$ or for list of values. (A1) for correct answer.[2 marks]

In (d) and (e) follow through from (c) if consistent wrong use of correct AP formula.

$110 + (n – 1) \times 10 > 176$ (A1)(M1)

$n = 8\therefore {\text{year 2007}}$ (A1)(A1)(ft)(G2)

Note: (A1) for $176$ or $66$ seen. (M1) for showing list of values and comparing them to $176$ or for equating formula to $176$ or for writing the inequality. If $n = 8$ not seen can still get (A2) for 2007. Answer $n = 8$ with no working gets (G1).

$110 + n \times 10 > 176$ (A1)(M1)

$n = 7\therefore {\text{year 2007}}$ (A1)(A1)(ft)(G2)[4 marks]

In (d) and (e) follow through from (c) if consistent wrong use of correct AP formula.

(i) $180$ (A1)(ft)

Grove School gets the money. (A1)(ft)

Note: (A1) for $180$ seen. (A1) for correct answer.

(ii) ${\text{100}} \times {\text{1}}{\text{.0}}{{\text{6}}^{n – 1}} > 110 + (n – 1) \times 10$ (M1)

$n = 20\therefore {\text{year 2019}}$ (A1)(A1)(ft)(G2)

Note: (M1) for showing lists of values for each school and comparing them or for equating both formulae or writing the correct inequality. If $n = 20$ not seen can still get (A2) for 2019. Follow through with ratio used in (b) and/or formula used in (d).

$100 \times {1.06^n} > 110 + n \times 10$ (M1)

$n = 19\therefore {\text{year 2019}}$ (A1)(A1)(ft)(G2)

graphically

Note: (M1) for sketch of both functions on the same graph, (A1) for the intersection point, (A1) for correct answer.[5 marks]

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

u_n = 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.

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.[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.

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.[2]

B.e.

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.[2]

B.f.

Answer/Explanation

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.

20r³= 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.

Question

A geometric sequence has $1024$ as its first term and $128$ as its fourth term.

Consider the arithmetic sequence $1{\text{, }}4{\text{, }}7{\text{, }}10{\text{, }}13{\text{, }} \ldots $

Show that the common ratio is $\frac{1}{2}$ .[2]

A.a.

Find the value of the eleventh term.[2]

A.b.

Find the sum of the first eight terms.[3]

A.c.

Find the number of terms in the sequence for which the sum first exceeds $2047.968$.[3]

A.d.

Find the value of the eleventh term.[2]

B.a.

The sum of the first $n$ terms of this sequence is $\frac{n}{2}(3n – 1)$.

(i) Find the sum of the first 100 terms in this arithmetic sequence.

(ii) The sum of the first $n$ terms is $477$.

(a) Show that $3{n^2} – n – 954 = 0$ .

(b) Using your graphic display calculator or otherwise, find the number of terms, $n$ .[6]

B.b.

Answer/Explanation

Markscheme

$1024{r^3} = 128$ (M1)

${r^3} = \frac{1}{8}$ or $r = \sqrt[3]{{\frac{1}{8}}}$ (M1)

$r = \frac{1}{2}{\text{ }}(0.5)$ (AG)

Notes: Award at most (M1)(M0) if last line not seen. Award (M1)(M0) if $128$ is found by repeated multiplication (division) of $1024$ by $0.5$ $(2)$.[2 marks]

A.a.

$1024 \times {0.5^{10}}$ (M1)

Notes: Award (M1) for correct substitution into correct formula. Accept an equivalent method.

1 (A1)(G2)[2 marks]

A.b.

${S_8} = \frac{{1024\left( {1 – {{\left( {\frac{1}{2}} \right)}^8}} \right)}}{{1 – \frac{1}{2}}}$ (M1)(A1)

Note: Award (M1) for substitution into the correct formula, (A1) for correct substitution.

(A1) for complete and correct list of eight terms (A1)

(M1) for their eight terms added (M1)

${S_8} = 2040$ (A1)(G2)[3 marks]

A.c.

$\frac{{1024\left( {1 – {{\left( {\frac{1}{2}} \right)}^n}} \right)}}{{1 – \frac{1}{2}}} > 2047.968$ (M1)(M1)(ft)

Notes: Award (M1) for correct substitution into the correct formula for the sum, (M1) for comparing to $2047.968$ . Accept equation. Follow through from their expression for the sum used in part (c).

If a list is used: ${S_{15}} = 2047.9375$ (M1)

${S_{16}} = 2047.96875$ (M1)

$n = 16$ (A1)(ft)(G2)

Note: Follow through from their expression for the sum used in part (c).[3 marks]

A.d.

${\text{common difference}} = 3$ (may be implied) (A1)

${u_{11}} = 31$ (A1)(G2)[2 marks]

B.a.

(i) $\frac{{100}}{2}(3 \times 100 – 1)$ OR $\frac{{100(2 + 99 \times 3)}}{2}$ (M1)

$14 950$ (A1)(G2)

(ii) (a) $\frac{n}{2}(3n – 1) = 477$ OR $\frac{n}{2}(2 + 3(n – 1)) = 477$ (M1)

$3{n^2} – n = 954$ (M1)

$3{n^2} – n – 954 = 0$ (AG)

Notes: Award second (M1) for correct removal of denominator or brackets and no further incorrect working seen. Award at most

(M1)(M0) if last line not seen.

(b) $18$ (G2)

Note: If both solutions to the quadratic equation are seen and the correct value is not identified as the required answer, award (G1)(G0).[6 marks]

B.b.

Question

Part A

The Green Park Amphitheatre was built in the form of a horseshoe and has 20 rows. The number of seats in each row increase by a fixed amount, d, compared to the number of seats in the previous row. The number of seats in the sixth row, u₆, is 100, and the number of seats in the tenth row, u₁₀, is 124. u₁ represents the number of seats in the first row.

Part B

Frank is at the amphitheatre and receives a text message at 12:00. Five minutes later he forwards the text message to three people. Five minutes later, those three people forward the text message to three new people. Assume this pattern continues and each time the text message is sent to people who have not received it before.

The number of new people who receive the text message forms a geometric sequence

1 , 3 , …

(i) Write an equation for u₆ in terms of d and u₁.

(ii) Write an equation for u₁₀ in terms of d and u₁.[2]

A.a.

Write down the value of

(i) d ;

(ii) u₁ .[2]

A.b.

Find the total number of seats in the amphitheatre.[3]

A.c.

A few years later, a second level was added to increase the amphitheatre’s capacity by another 1600 seats. Each row has four more seats than the previous row. The first row on this level has 70 seats.

Find the number of rows on the second level of the amphitheatre.[4]

A.d.

Write down the next two terms of this geometric sequence.[1]

B.a.

Write down the common ratio of this geometric sequence.[1]

B.b.

Calculate the number of people who will receive the text message at 12:30.[2]

B.c.

Calculate the total number of people who will have received the text message by 12:30.[2]

B.d.

Calculate the exact time at which a total of 29 524 people will have received the text message.[3]

B.e.

Answer/Explanation

Markscheme

(i) u₁ + 5d = 100 (A1)

(ii) u₁ + 9d = 124 (A1)[2 marks]

A.a.

(i) 6 (G1)(ft)

(ii) 70 (G1)(ft)

Notes: Follow through from their equations in parts (a) and (b) even if working not seen. Their answers must be integers. Award (M1)(A0) for an attempt to solve two equations analytically.[2 marks]

A.b.

$S_{20} = \frac{20}{2}(2 \times 70 + (20 – 1) \times 6)$ (M1)(A1)(ft)

Note: Award (M1) for substituted sum of AP formula, (A1)(ft) for their correct substituted values.

= 2540 (A1)(ft)(G2)

Note: Follow through from their part (b).[3 marks]

A.c.

$\frac{n}{2}(2 \times 70 + (n – 1) \times 4) = 1600$ (M1)(A1)

Note: Award (M1) for substituted sum of AP formula, (A1) for their correct substituted values.

4n² +136n – 3200 = 0 (M1)

Note: Award (M1) for this equation (or other equivalent expanded quadratic) seen, may be implied if correct final answer seen.

n = 16 (A1)(G3)

Note: Do not award the final (A1) for n = 16, – 50 given as final answer, award (G2) if n = 16, – 50 given as final answer without working.[4 marks]

A.d.

9, 27 (A1)[1 mark]

B.a.

3 (A1)[1 mark]

B.b.

1 × 3⁶ (M1)

= 729 (A1)(ft)(G2)

Note: Award (M1) for correctly substituted GP formula. Follow through from their answer to part (b).[2 marks]

B.c.

$\frac{{1(3^7 – 1)}}{(3 – 1)}$ (M1)

Note: Award (M1) for correctly substituted GP formula. Accept sum 1+ 3 + 9 + 27 + … + 729. If lists are used, award (M1) for correct list that includes 1093. (1, 4, 13, 40, 121, 364, 1093, 3280…)

= 1093 (A1)(ft)(G2)

Note: Follow through from their answer to part (b). For consistent use of n = 6 from part (c) (243) to part (d) leading
to an answer of 364, treat as double penalty and award (M1)(A1)(ft) if working is shown.[2 marks]

B.d.

$\frac{{1(3^n – 1)}}{(3 – 1)} = 29524$ (M1)

Note: Award (M1) for correctly substituted GP formula. If lists are used, award (M1) for correct list that includes 29524. (1, 4, 13, 40, 121, 364, 1093, 3280, 9841, 29524, 88573…). Accept alternative methods, for example continuation of sum in part (d).

n = 10 (A1)(ft)

Note: Follow through from their answer to part (b).

Exact time = 12:45 (A1)(ft)(G2)[3 marks]

B.e.

Question

On Monday Paco goes to a running track to train. He runs the first lap of the track in 120 seconds. Each lap Paco runs takes him 10 seconds longer than his previous lap.

Find the time, in seconds, Paco takes to run his fifth lap.[3]

Paco runs his last lap in 260 seconds.

Find how many laps he has run on Monday.[3]

Find the total time, in minutes, run by Paco on Monday.[4]

On Wednesday Paco takes Lola to train. They both run the first lap of the track in 120 seconds. Each lap Lola runs takes 1.06 times as long as her previous lap.

Find the time, in seconds, Lola takes to run her third lap.[3]

Find the total time, in seconds, Lola takes to run her first four laps.[3]

Each lap Paco runs again takes him 10 seconds longer than his previous lap. After a certain number of laps Paco takes less time per lap than Lola.

Find the number of the lap when this happens.[3]

Answer/Explanation

Markscheme

$120 + 10 \times 4$ (M1)(A1)

Notes: Award (M1) for substituted AP formula, (A1) for correct substitutions. Accept a list of 4 correct terms.

= 160 (A1)(G3)

$120 + (n – 1) \times 10 = 260$ (M1)(M1)

Notes: Award (M1) for correctly substituted AP formula, (M1) for equating to 260. Accept a list of correct terms showing at least the 14^th and 15^th terms.

= 15 (A1)(G2)

$\frac{{15}}{2}(120 + 260)$ or $\frac{{15}}{2}(2 \times 120 + (15 – 1) \times 10)$ (M1)(A1)(ft)

Notes: Award (M1) for substituted AP sum formula, (A1)(ft) for correct substitutions. Accept a sum of a list of 15 correct terms. Follow through from their answer to part (b).

2850 seconds (A1)(ft)(G2)

Note: Award (G2) for 2850 seen with no working shown.

47.5 minutes (A1)(ft)(G3)

Notes: A final (A1)(ft) can be awarded for correct conversion from seconds into minutes of their incorrect answer. Follow through from their answer to part (b).

$120 \times {1.06^{3 – 1}}$ (M1)(A1)

Notes: Award (M1) for substituted GP formula, (A1) for correct substitutions. Accept a list of 3 correct terms.

= 135 (134.832) (A1)(G2)

${S_4} = \frac{{120({{1.06}^4} – 1)}}{{(1.06 – 1)}}$ (M1)(A1)

Notes: Award (M1) for substituted GP sum formula, (A1) for correct substitutions. Accept a sum of a list of 4 correct terms.

= 525 (524.953…) (A1)(G2)

$120 + (n – 1) \times 10 < 120 \times {1.06^{n – 1}}$ (M1)(M1)

Notes: Award (M1) for correct left hand side, (M1) for correct right hand side. Accept an equation. Follow through from their expressions given in parts (a) and (d).

List of at least 2 terms for both sequences (120, 130, … and 120, 127.2, …) (M1)

List of correct 12^th and 13^th terms for both sequences (…, 230, 240 and …, 227.8, 241.5) (M1)

A sketch with a line and an exponential curve, (M1)

An indication of the correct intersection point (M1)

13^th lap (A1)(ft)(G2)

Note: Do not award the final (A1)(ft) if final answer is not a positive integer.

Question

Consider the sequence ${u_1},{\text{ }}{u_2},{\text{ }}{u_3},{\text{ }} \ldots ,{\text{ }}{{\text{u}}_n},{\text{ }} \ldots $ where

\[{u_1} = 600,{\text{ }}{u_2} = 617,{\text{ }}{u_3} = 634,{\text{ }}{u_4} = 651.\]

The sequence continues in the same manner.

Find the value of ${u_{20}}$.[3]

Find the sum of the first 10 terms of the sequence.[3]

Now consider the sequence ${v_1},{\text{ }}{v_2},{\text{ }}{v_3},{\text{ }} \ldots ,{\text{ }}{v_n},{\text{ }} \ldots $ where

\[{v_1} = 3,{\text{ }}{v_2} = 6,{\text{ }}{v_3} = 12,{\text{ }}{v_4} = 24\]

This sequence continues in the same manner.

Find the exact value of ${v_{10}}$.[3]

Now consider the sequence ${v_1},{\text{ }}{v_2},{\text{ }}{v_3},{\text{ }} \ldots ,{\text{ }}{v_n},{\text{ }} \ldots $ where

\[{v_1} = 3,{\text{ }}{v_2} = 6,{\text{ }}{v_3} = 12,{\text{ }}{v_4} = 24\]

This sequence continues in the same manner.

Find the sum of the first 8 terms of this sequence.[3]

$k$ is the smallest value of $n$ for which ${v_n}$ is greater than ${u_n}$.

Calculate the value of $k$.[3]

Answer/Explanation

Markscheme

$600 + (20 – 1) \times 17$ (M1)(A1)

Note: Award (M1) for substituted arithmetic sequence formula, (A1) for correct substitutions. If a list is used, award (M1) for at least 6 correct terms seen, award (A1) for at least 20 correct terms seen.

$ = 923$ (A1)(G3)[3 marks]

$\frac{{10}}{2}\left[ {2 \times 600 + (10 – 1) \times 17} \right]$ (M1)(A1)

Note: Award (M1) for substituted arithmetic series formula, (A1) for their correct substitutions. Follow through from part (a). For consistent use of geometric series formula in part (b) with the geometric sequence formula in part (a) award a maximum of (M1)(A1)(A0) since their final answer cannot be an integer.

${u_{10}} = 600 + (10 – 1)17 = 753$ (M1)

${S_{10}} = \frac{{10}}{2}\left( {600 + {\text{their }}{u_{10}}} \right)$ (M1)

Note: Award (M1) for their correctly substituted arithmetic sequence formula, (M1) for their correctly substituted arithmetic series formula. Follow through from part (a) and within part (b).

Note: If a list is used, award (M1) for at least 10 correct terms seen, award (A1) for these terms being added.

$ = 6765$ (accept $6770$) (A1)(ft)(G2)[3 marks]

$3 \times {2^9}$ (M1)(A1)

Note: Award (M1) for substituted geometric sequence formula, (A1) for correct substitutions. If a list is used, award (M1) for at least 6 correct terms seen, award (A1) for at least 8 correct terms seen.

$ = 1536$ (A1)(G3)

Note: Exact answer only. If both exact and rounded answer seen, award the final (A1).[3 marks]

$\frac{{3 \times \left( {{2^8} – 1} \right)}}{{2 – 1}}$ (M1)(A1)(ft)

Note: Award (M1) for substituted geometric series formula, (A1) for their correct substitutions. Follow through from part (c). If a list is used, award (M1) for at least 8 correct terms seen, award (A1) for these 8 correct terms being added. For consistent use of arithmetic series formula in part (d) with the arithmetic sequence formula in part (c) award a maximum of (M1)(A1)(A1).

$ = 765$ (A1)(ft)(G2)[3 marks]

$3 \times {2^{k – 1}} > 600 + (k – 1)(17)$ (M1)

Note: Award (M1) for their correct inequality; allow equation.

Follow through from parts (a) and (c). Accept sketches of the two functions as a valid method.

$k > 8.93648 \ldots $ (may be implied) (A1)(ft)

Note: Award (A1) for $8.93648…$ seen. The GDC gives answers of $-34.3$ and $8.936$ to the inequality; award (M1)(A1) if these are seen with working shown.

${v_8} = 384$ ${u_8} = 719$ (M1)

${v_9} = 768$ ${u_9} = 736$ (M1)

Note: Award (M1) for ${v_8}$ and ${u_8}$ both seen, (M1) for ${v_9}$ and ${u_9}$ both seen.

$k = 9$ (A1)(ft)(G2)

Note: Award (G1) for $8.93648…$ and $-34.3$ seen as final answer without working. Accept use of $n$.[3 marks]

Question

Give your answers to parts (a) to (e) to the nearest dollar.

On Hugh’s 18th birthday his parents gave him options of how he might receive his monthly allowance for the next two years.

Option A $\$60$ each month for two years

Option B $\$10$ in the first month, $\$15$ in the second month, $\$20$ in the third month, increasing by $\$5$ each month for two years

Option C $\$15$ in the first month and increasing by $10\%$ each month for two years

Option D Investing $\$1500$ at a bank at the beginning of the first year, with an interest rate of $6\%$ per annum, compounded monthly.

Hugh does not spend any of his allowance during the two year period.

If Hugh chooses Option A, calculate the total value of his allowance at the end of the two year period.[2]

If Hugh chooses Option B, calculate

(i) the amount of money he will receive in the 17th month;

(ii) the total value of his allowance at the end of the two year period.[5]

If Hugh chooses Option C, calculate

(i) the amount of money Hugh would receive in the 13th month;

(ii) the total value of his allowance at the end of the two year period.[5]

If Hugh chooses Option D, calculate the total value of his allowance at the end of the two year period.[3]

State which of the options, A, B, C or D, Hugh should choose to give him the greatest total value of his allowance at the end of the two year period.[1]

Another bank guarantees Hugh an amount of $\$1750$ after two years of investment if he invests $1500 at this bank. The interest is compounded annually.

Calculate the interest rate per annum offered by the bank.[3]

Answer/Explanation

Markscheme

The first time an answer is not given to the nearest dollar in parts (a) to (e), the final (A1) in that part is not awarded.

$60 \times 24$ (M1)

Note: Award (M1) for correct product.

$ = 1440$ (A1)(G2)[2 marks]

The first time an answer is not given to the nearest dollar in parts (a) to (e), the final (A1) in that part is not awarded.

(i) $10 + (17 – 1)(5)$ (M1)(A1)

Note: Award (M1) for substituted arithmetic sequence formula, (A1) for correct substitution.

$ = 90$ (A1)(G2)

(ii) $\frac{{24}}{2}\left( {2(10) + (24 – 1)(5)} \right)$ (M1)

$\frac{{24}}{2}\left( {10 + 125} \right)$ (M1)

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

$ = 1620$ (A1)(ft)(G1)

Note: Follow through from part (b)(i).[5 marks]

The first time an answer is not given to the nearest dollar in parts (a) to (e), the final (A1) in that part is not awarded.

(i) $15{(1.1)^{12}}$ (M1)(A1)

Note: Award (M1) for substituted geometric sequence formula, (A1) for correct substitutions.

$ = 47$ (A1)(G2)

Note: Award (M1)(A1)(A0) for $47.08$.

Award (G1) for $47.08$ if workings are not shown.

(ii) $\frac{{15({{1.1}^{24}} – 1)}}{{1.1 – 1}}$ (M1)

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

$ = 1327$ (A1)(ft)(G1)

Note: Follow through from part (c)(i).[5 marks]

The first time an answer is not given to the nearest dollar in parts (a) to (e), the final (A1) in that part is not awarded.

$1500{\left( {1 + \frac{6}{{100(12)}}} \right)^{12(2)}}$ (M1)(A1)

Note: Award (M1) for substituted compound interest formula, (A1) for correct substitutions.

$N = 2$

$I\% = 6$

$PV = 1500$

$P/Y = 1$

$C/Y = 12$ (A1)(M1)

Note: Award (A1) for $C/Y = 12$ seen, (M1) for other correct entries.

$N = 24$

$I\% = 6$

$PV = 1500$

$P/Y = 12$

$C/Y = 12$ (A1)(M1)

Note: Award (A1) for $C/Y = 12$ seen, (M1) for other correct entries.

$ = 1691$ (A1)(G2)[3 marks]

The first time an answer is not given to the nearest dollar in parts (a) to (e), the final (A1) in that part is not awarded.

Option D (A1)(ft)

Note: Follow through from their parts (a), (b), (c) and (d). Award (A1)(ft) only if values for the four options are seen and only if their answer is consistent with their parts (a), (b), (c) and (d).[1 mark]

$1750 = 1500{\left( {1 + \frac{r}{{100}}} \right)^2}$ (M1)(A1)

Note: Award (M1) for substituted compound interest formula equated to $1750$, (A1) for correct substitutions into formula.

$N = 2$

$PV = 1500$

$FV = – 1750$

$P/Y = 1$

$C/Y = 1$ (A1)(M1)

Note: Award (A1) for $FV = 1750$ seen, (M1) for other correct entries.

$ = 8.01\% {\text{ (8.01234}} \ldots \% ,{\text{ }}0.0801{\text{)}}$ (A1)(G2)[3 marks]

Question

In a game, n small pumpkins are placed 1 metre apart in a straight line. Players start 3 metres before the first pumpkin.

Each player collects a single pumpkin by picking it up and bringing it back to the start. The nearest pumpkin is collected first. The player then collects the next nearest pumpkin and the game continues in this way until the signal is given for the end.

Sirma runs to get each pumpkin and brings it back to the start.

Write down the distance, ${a_1}$, in metres that she has to run in order to collect the first pumpkin.[1]

The distances she runs to collect each pumpkin form a sequence ${a_1},{\text{ }}{a_2},{\text{ }}{a_3}, \ldots $ .

(i) Find ${a_2}$.

(ii) Find ${a_3}$.[2]

Write down the common difference, $d$, of the sequence.[1]

The final pumpkin Sirma collected was 24 metres from the start.

(i) Find the total number of pumpkins that Sirma collected.

(ii) Find the total distance that Sirma ran to collect these pumpkins.[5]

Peter also plays the game. When the signal is given for the end of the game he has run 940 metres.

Calculate the total number of pumpkins that Peter collected.[3]

Peter also plays the game. When the signal is given for the end of the game he has run 940 metres.

Calculate Peter’s distance from the start when the signal is given.[2]

Answer/Explanation

Markscheme

$6{\text{ (m)}}$ (A1)(G1)

(i) $8$ (A1)(ft)

(ii) $10$ (A1)(ft)(G2)

Note: Follow through from part (a).

$2{\text{ (m)}}$ (A1)(ft)

Note: Follow through from parts (a) and (b).

(i) $2 \times 24 = 6 + 2(n – 1)\;\;\;$OR$\;\;\;24 = 3 + (n – 1)$ (M1)

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

$n = 22$ (A1)(ft)(G1)

Note: Follow through from parts (a) and (c).

(ii) $\frac{{(6 + 48)}}{2} \times 22$ (M1)(A1)(ft)

Note: Award (M1) for substitution in arithmetic series formula, (A1)(ft) for correct substitution.

$ = 594$ (A1)(ft)(G2)

Note: Follow through from parts (a) and (d)(i).

$\frac{{\left[ {2 \times 6 + 2(n – 1)} \right] \times n}}{2} = 940$ (M1)(A1)(ft)

Notes: Award (M1) for substitution in arithmetic series formula, (A1) for their correct substituted formula equated to $940$. Follow through from parts (a) and (c).

${n^2} + 5n – 940 = 0$

$n = 28.2611 \ldots $

$n = 28$ (A1)(ft)(G2)

$\frac{{\left[ {2 \times 6 + 2(28 – 1)} \right] \times 28}}{2}$ (M1)

Notes: Award (M1) for substituting their $28$ into the arithmetic series formula.

$ = 16{\text{ (m)}}$ (A1)(ft)(G2)

Question

The sum of the first $n$ terms of an arithmetic sequence is given by ${S_n} = 6n + {n^2}$.

Write down the value of

(i) ${S_1}$;

(ii) ${S_2}$.[2]