IB DP Math AI: Topic : SL 1.3: Geometric sequences and series: IB style Questions HL Paper 2

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.

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

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

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

ii.b.A second geometric progression G2 has the form \(1,\frac{1}{3},\frac{1}{9},\frac{1}{{27}}…\)[1]

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

ii.d.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.e.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]

▶️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.

a.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]

b.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]

c.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]

d.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 \( = \frac{{20}}{2}(2500 + 6300)\)

OR

\( \frac{{20}}{2}(5000 + 19 \times 200)\)     (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]

a.

(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]

b.

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]

c.

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]

d.

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.

a.Find the number of students attending Park School in

(i)     January 2001;

(ii)    January 2003.[4]

b.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]

c.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]

d.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]

e.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]

a.

\(100 \times {1.06^7} = 150.36 \ldots  = 150\) 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]

b.

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

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

c.

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).

OR

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

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

d.

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).

OR

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

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

OR

graphically

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

e.

Question

A geometric sequence has second term 12 and fifth term 324.

Consider the following propositions

p: The number is a multiple of five.

q: The number is even.

r: The number ends in zero.

i, a.Calculate the value of the common ratio.[4]

i, b.Calculate the 10th term of this sequence.[3]

i, c.The kth term is the first term which is greater than 2000. Find the value of k.[3]

ii, a.Write in words \((q \wedge \neg r) \Rightarrow \neg p\).[3]

ii, b, i.Consider the statement “If the number is a multiple of five, and is not even then it will not end in zero”.

Write this statement in symbolic form.[4]

ii, b, ii.Consider the statement “If the number is a multiple of five, and is not even then it will not end in zero”.

Write the contrapositive of this statement in symbolic form.[2]

▶️Answer/Explanation

Markscheme

u1r4 = 324     (A1)

u1r = 12     (A1)

r3 = 27     (M1)

r = 3     (A1)(G3)

Note: Award at most (G3) for trial and error.[4 marks]

i, a.

4 × 39 = 78732 or 12 × 38 = 78732     (A1)(M1)(A1)(ft)(G3)

Note: Award (A1) for u1 = 4 if n = 9 , or u1 = 12 if n = 8, (M1) for correctly substituted formula.

(ft) from their (a).[3 marks]

i, b.

4 × 3k−1 > 2000     (M1)

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

k > 6     (A1)

k = 7     (A1)(ft)(G2)

Notes: If second line not seen award (A2) for correct answer. (ft) from their (a).

Accept a list, must see at least 3 terms including the 6th and 7th.

Note: If arithmetic sequence formula is used consistently in parts (a), (b) and (c), award (A0)(A0)(M0)(A0) for (a) and (ft) for parts (b) and (c).[3 marks]

i, c.

If the number is even and the number does not end in zero, (then) the number is not a multiple of five.     (A1)(A1)(A1)

Note: Award (A1) for “if…(then)”, (A1) for “the number is even and the number does not end in zero”, (A1) for the number is not a multiple of 5.[3 marks]

ii, a.

\((p \wedge \neg q) \Rightarrow \neg r\)     (A1)(A1)(A1)(A1)

(A1) for \(\Rightarrow\), (A1) for \(\wedge\), (A1) for p and \(\neg q\), (A1) for \(\neg r\)

Note: If parentheses not present award at most (A1)(A1)(A1)(A0).[4 marks]

ii, b, i.

\(r \Rightarrow (\neg p \vee q)\)   OR   \(r \Rightarrow \neg (p \wedge \neg q)\)     (A1)(ft)(A1)(ft)

Note: Award (A1)(ft) for reversing the order, (A1) for negating the statements on both sides.

If parentheses not present award at most (A1)(ft)(A0).

Do not penalise twice for missing parentheses in (i) and (ii).[2 marks]

ii, b, ii.

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, …

A.a.Calculate the value of his investment at the end of the third year for each investment option, correct to two decimal places.[8]

A.b.Determine Daniel’s best investment option.[1]

B.a.Calculate u1, the first term in the sequence.[2]

B.b.Show that the common difference is 7.[2]

B.c.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.d.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.e.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.f.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]

▶️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.

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.
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