# IBDP Maths AI: Topic : SL 1.3: Geometric sequences and series: IB style Questions SL Paper 1

## Question

The first term of an arithmetic sequence is $$0$$ and the common difference is $$12$$.

Find the value of the $${96^{{\text{th}}}}$$ term of the sequence.

a.

The first term of a geometric sequence is $$6$$. The $${6^{{\text{th}}}}$$ term of the geometric sequence is equal to the $${17^{{\text{th}}}}$$ term of the arithmetic sequence given above.

Write down an equation using this information.

b.

The first term of a geometric sequence is $$6$$. The \$${6^{{\text{th}}}}$$ term of the geometric sequence is equal to the $${17^{{\text{th}}}}$$ term of the arithmetic sequence given above.

Calculate the common ratio of the geometric sequence.

c.

## Markscheme

$${u_{96}} = {u_1} + 95d$$     (M1)

$$= 0 + 95 \times 12$$

$$= 1140$$     (A1)     (C2)

[2 marks]

a.

$$6{r^5} = 16d$$     (A1)

$$6{r^5} = 16 \times 12$$ ($$192$$)     (A1)     (C2)

Note: (A1) only, if both terms seen without an equation.[2 marks]

b.

$${r^5} = 32$$     (A1)(ft)

Note: (ft) from their (b).

$$r = 2$$     (A1)(ft)     (C2)[2 marks]

c.

## Question

The annual fees paid to a school for the school years 2000, 2001 and 2002 increase as a geometric progression. The table below shows the fee structure. Calculate the common ratio for the increasing sequence of fees.

a.

The fees continue to increase in the same ratio.

Find the fees paid for 2006.

b.

The fees continue to increase in the same ratio.

A student attends the school for eight years, starting in 2000.

Find the total fees paid for these eight years.

c.

## Markscheme

$$r = \frac{{8320}}{{8000}}$$ (or equivalent)     (M1)

Note: Award (M1) for dividing correct terms.

r = 1.04     (A1)     (C2)

Notes: In (b) and (c) (ft) from candidate’s r.

Allow lists, graphs etc. as working in (b) and (c).[2 marks]

a.

Financial penalty (FP) applies in this part

Fees = 8000 (1.04)6     (M1)

Note: Award (M1) for correct substitution into correct formula.(FP)     Fees = 10122.55 USD (USD not required)     (A1)(ft)     (C2)

Note: Special exception to the note above.

Award maximum of (M1)(A0) if 5 is used as the power.[2 marks]

b.

Financial penalty (FP) applies in this part

$${\text{Total}} = \frac{{8000({{1.04}^8} – 1)}}{{1.04 – 1}}$$     (M1)

Notes: Award (M1) for correct substitution into correct formula.

Give full credit for solution by lists.

(FP)     Total = 73713.81 USD (USD not required)     (A1)(ft)     (C2)[2 marks]

c.

## Question

The population of big cats in Africa is increasing at a rate of 5 % per year. At the beginning of 2004 the population was $$10\,000$$.

Write down the population of big cats at the beginning of 2005.

a.

Find the population of big cats at the beginning of 2010.

b.

Find the number of years, from the beginning of 2004, it will take the population of big cats to exceed $$50\,000$$.

c.

## Markscheme

$$10\,000 \times 1.05$$

$$= 10\,500$$     (A1)     (C1)[1 mark]

a.

$$10\,000 \times {1.05^6}$$     (M1)

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

$$= 13\,400$$     (A1)     (C2)[2 marks]

b.

$$50\,000 = 10\,000 \times 1.05”$$     (M1)(A1)

Note: Award (M1) for $$10\,000 \times 1.05”$$ or equivalent, (A1) for $$50\,000$$

$$n = 33.0$$ (Accept 33)     (A1)     (C3)[3 marks]

c.

## Question

Consider the geometric sequence 16, 8, a, 2, b, …

Write down the common ratio.

a.

Write down the value of a.

b.i.

Write down the value of b.

b.ii.

The sum of the first n terms is 31.9375. Find the value of n.

c.

## Markscheme

0.5     $$\left( {\frac{1}{2}} \right)$$     (A1)    (C1)[1 mark]

a.

4     (A1)[1 mark]

b.i.

1     (A1)     (C2)[1 mark]

b.ii.

$$\frac{{16(1 – {{0.5}^n})}}{{(1 – 0.5)}} = 31.9375$$     (M1)(M1)

Note: Award (M1) for correct substitution in the GP formula, (M1) for equating their sum to 31.9375.

OR

sketch of the function  $$y = \frac{{16(1 – {{0.5}^n})}}{{(1 – 0.5)}}$$     (M1)

indication of point where y = 31.9375     (M1)

OR

16 + 8 + 4 + 2 + 1 + 0.5 + 0.25 + 0.125 + 0.0625 = 31.9375     (M1)(M1)

Note: Award (M1) for a list of at least 7 correct terms, (A1) for the sum of the terms equated to 31.9375.

n = 9     (A1)(ft)     (C3)

Note: Follow through from their answer to part (a) but answer mark is lost if n is not a whole number.[3 marks]

c.

## Question

The seventh term, $${u_7}$$ , of a geometric sequence is $$108$$. The eighth term, $${u_8}$$ , of the sequence is $$36$$ .

Write down the common ratio of the sequence.

a.

Find $${u_1}$$ .

b.

The sum of the first $$k$$ terms in the sequence is $$118 096$$ . Find the value of $$k$$ .

c.

## Markscheme

$$r = \frac{{36}}{{108}}\left( {\frac{1}{3}} \right)$$     (A1)     (C1)

Note: Accept $$0.333$$.[1 mark]

a.

$${u_1}{\left( {\frac{1}{3}} \right)^7} = 36$$     (M1)

Note: Award (M1) for correct substitution in formula for nth term of a GP. Accept equivalent forms.

$${u_1} = 78732$$     (A1)(ft)     (C2)

Notes: Accept $$78 700$$. Follow through from their common ratio found in part (a). If $$0.333$$ used from part (a) award (M1)(A1)(ft) for an answer of $$79 285$$ or $$79 300$$ irrespective of whether working is shown.[2 marks]

b.

$$118096 = \frac{{78732\left( {1 – {{\left( {\frac{1}{3}} \right)}^k}} \right)}}{{\left( {1 – \frac{1}{3}} \right)}}$$     (M1)(M1)

Notes: Award (M1) for correct substitution in the sum of a GP formula, (M1) for equating their sum to $$118 096$$. Follow through from parts (a) and (b).

OR

Sketch of the function $$y = 78732\frac{{\left( {1 – {{\left( {\frac{1}{3}} \right)}^k}} \right)}}{{\left( {1 – \frac{1}{3}} \right)}}$$     (M1)
Indication of point where $$y = 118 096$$     (M1)

OR

$$78 732 + 26 244 + 8748 + 2916 + 972 + 324 + 108 + 36 + 12 + 4 = 118 096$$     (M1)(M1)

Note: Award (M1) for a list of at least 8 correct terms, (M1) for the sum of the terms equated to $$118 096$$.

$$k =10$$     (A1)(ft)     (C3)

Notes: Follow through from parts (a) and (b). If k is not an integer, do not award final (A1). Accept alternative methods. If $$0.333$$ and $$79 285$$ used award (M1)(M1)(A1)(ft) for $$k = 5$$. If $$0.333$$ and $$79 300$$ used award (M1)(M1)(A0).[3 marks]

c.

## Question

Consider the sequence

${\text{512, 128, 32, 8, }} \ldots$

Calculate the exact value of the ninth term of the sequence.

a.

Calculate the least number of terms required for the sum of the sequence to be greater than 682.6

b.

## Markscheme

$${u_9} = 512{\left( {\frac{1}{4}} \right)^8}$$     (M1)(A1)

Notes: Award (M1) for substituted geometric sequence formula, (A1) for correct substitution.

OR

If a list is used, award (M1) for a list of 9 terms, (A1) for all 9 terms correct.     (M1)(A1)

$$= \frac{1}{{128}}$$ ($$0.0078125$$)     (A1)     (C3)

Note: Award (A1) for exact answer only.[3 marks]

a.

$$\frac{{512\left( {1 – {{\left( {\frac{1}{4}} \right)}^n}} \right)}}{{1 – \left( {\frac{1}{4}} \right)}} > 682.6$$     (M1)(A1)(ft)

Notes: Award (M1) for setting substituted geometric sum formula $$> 682.6$$ (A1)(ft) for correct substitution into geometric sum formula. Follow through from their common ratio.

OR

If list is used, award (M1) for S(6) and S(7) seen, values don’t have to be correct.

(A1) for correct S(6) and S(7). (S(6) $$= 682.5$$ and S(7) $$= 682.625$$).     (M1)(A1)

$$n = 7$$     (A1)(ft)     (C3)

Notes: Follow through from their common ratio. Do not award the final (A1)(ft) if $$n$$ is less than $$5$$ or if $$n$$ is not an integer.[3 marks]

b.

## Question

Javier starts training for a running race.

On the first day he runs 1.5 km. Every day he runs 10 % more than the day before.

Write down the distance he runs on the second day of training.

a.

Calculate the total distance Javier runs in the first seven days of training.

b.

Javier stops training on the day his total distance exceeds 100 km.

Calculate the number of days Javier has trained for the running race.

c.

## Markscheme

1.65 (km) or 1650 (m)     (A1)     (C1)[1 mark]

a.

$$\frac{{1.5({{1.1}^7} – 1)}}{{1.1 – 1}}$$     (M1)

Notes: Award (M1) for correct substitution of candidate’s 10 % into the correct formula. Accept a list.

14.2 (km)     (A1)(ft)     (C2)[2 marks]

b.

$$\frac{{1.5({{1.1}^n} – 1)}}{{1.1 – 1}} > 100$$     (M1)

Note: Award (M1) for setting up their inequality/equation. Accept a list.

n = 21.371…     (A1)(ft)

n = 22     (A1)(ft)     (C3)

Notes: Follow through from their values of 1.1 and 1.5 in part (b). The final (A1)(ft) is for rounding up their answer for n to a whole number of days.[3 marks]

c.

## Question

The first term of a geometric sequence is 2 and the third term is 2.205.

Calculate the common ratio of the sequence;

a.

Calculate the eleventh term of the sequence;

b.

Calculate the sum of the first 23 terms of the sequence.

c.

## Markscheme

2r2 = 2.205     (M1)

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

r = 1.05     (A1)     (C2)[2 marks]

a.

2(1.05)10     (M1)

Note: Award (M1) for the correct substitution, using their answer to part (a), in geometric sequence formula.

= 3.26     (3.25778…)     (A1)(ft)     (C2)

Note: Follow through from their part (a).[2 marks]

b.

$$\frac{{2({{1.05}^{23}} – 1)}}{{(1.05 – 1)}}$$     (M1)

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

= 82.9     (82.8609…)     (A1)(ft)     (C2)

Notes: Accept an answer of 3.97221…if r = −1.05 is found in part (a) and used again in part (c). Follow through from their part (a).[2 marks]

c.

## Question

The fourth term, u4, of a geometric sequence is 135. The fifth term, u5, is 101.25 .

Find the common ratio of the sequence.

a.

Find u1, the first term of the sequence.

b.

Calculate the sum of the first 10 terms of the sequence.

c.

## Markscheme

$$\frac{{101.25}}{{135}}$$     (M1)

$$= \frac{3}{4}(0.75)$$     (A1)     (C2)

a.

$${u_1}{\left( {\frac{3}{4}} \right)^4} = 101.25$$     (M1)

OR

$${u_1}{\left( {\frac{3}{4}} \right)^3} = 135$$     (M1)

OR

(by list)

$${u_3} = 180,{\text{ }}{u_2} = 240$$     (M1)

Notes: Award (M1) for their correct substitution in geometric sequence formula, or stating explicitly $${u_3}$$ and $${u_2}$$.

$$({u_1} = )320$$     (A1)(ft)     (C2)

b.

$${S_{10}} = \frac{{320\left( {1 – {{\left( {\frac{3}{4}} \right)}^{10}}} \right)}}{{1 – \left( {\frac{3}{4}} \right)}}$$     (M1)

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

Accept a list of all their ten geometric terms.

= 1210 (1207.918…)     (A1)(ft)     (C2)

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

c.

## Question

$$512$$ competitors enter round 1 of a tennis tournament, in which each competitor plays a match against one other competitor.

The winning competitor progresses to the next round (round 2); the losing competitor leaves the tournament.

The tournament continues in this manner until there is a winner.

Find the number of competitors who play in round 6 of the tournament.

a.

Find the total number of matches played in the tournament.

b.

## Markscheme

$$512{\left( {\frac{1}{2}} \right)^5}$$     (M1)(A1)

Note: Award (M1) for substituted geometric progression formula, (A1) for correct substitution.

If a list is used, award (M1) for a list of at least six terms, beginning with $$512$$ and (A1) for first six terms correct.

$$16$$     (A1)     (C3)[3 marks]

a.

$${S_9} = 256\left( {\frac{{1 – {{\left( {\frac{1}{2}} \right)}^9}}}{{1 – \frac{1}{2}}}} \right)$$   OR   $$\frac{{({2^9} – 1)}}{{2 – 1}}$$     (M1)(A1)

Note: Award (M1) for substituted sum of a GP formula, (A1) for correct substitution.

If a list is used, award (A1) for at least 9 correct terms, including $$1$$, and (M1) for their 9 terms, including $$1$$, added together.

$$511$$     (A1)     (C3)[3 marks]

b.

## Question

The second term of an arithmetic sequence is 30. The fifth term is 90.

Calculate

(i)     the common difference of the sequence;

(ii)     the first term of the sequence.

a.

The first, second and fifth terms of this arithmetic sequence are the first three terms of a geometric sequence.

Calculate the seventh term of the geometric sequence.

b.

## Markscheme

(i)     $${u_1} + d = 30,{\text{ }}{u_1} + 4d = 90,{\text{ }}3d = 90 – 30\;\;\;$$(or equivalent)     (M1)

Note: Award (M1) for one correct equation. Accept a list of at least 5 correct terms.

$$(d = ){\text{ }}20$$     (A1)

(ii)     $$({u_1} = ){\text{ }}10$$     (A1)(ft)     (C3)

Note: Follow through from (a)(i), irrespective of working shown if $${u_1} = 30 – {\text{ (their }}d)\;\;\;$$OR$$\;\;\;{u_1} = 90 – 4 \times {\text{ (their }}d{\text{)}}$$

a.

$$({u_7} = ){\text{ }}10({3^{(7 – 1)}}\;\;\;$$OR$$\;\;\;({u_7} = ){\text{ 10}} \times {3^6}$$     (M1)(A1)(ft)

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

OR

$$10;{\text{ }}30;{\text{ }}90;{\text{ }}270;{\text{ }}810;{\text{ }}2430;{\text{ }}7290$$     (M1)(A1)(ft)

Note: Award (M1) for a list of at least 5 consecutive terms of a geometric sequence, (A1)(ft) for terms corresponding to their answers in part (a).

$$= 7290$$     (A1)(ft)     (C3)

Note: Follow through from part (a).

b.

## Question

Only one of the following four sequences is arithmetic and only one of them is geometric.

$${a_n} = 1,{\text{ }}2,{\text{ }}3,{\text{ }}5,{\text{ }} \ldots$$

$${b_n} = 1,{\text{ }}\frac{3}{2},{\text{ }}\frac{9}{4},{\text{ }}\frac{{27}}{8},{\text{ }} \ldots$$

$${c_n} = 1,{\text{ }}\frac{1}{2},{\text{ }}\frac{1}{3},{\text{ }}\frac{1}{4},{\text{ }} \ldots$$

$${d_n} = 1,{\text{ }}0.95,{\text{ }}0.90,{\text{ }}0.85,{\text{ }} \ldots$$

State which sequence is

(i)     arithmetic;

(ii)     geometric.

a.

For another geometric sequence $${e_n} = – 6,{\text{ }} – 3,{\text{ }} – \frac{3}{2},{\text{ }} – \frac{3}{4},{\text{ }} \ldots$$

write down the common ratio;

b(i).

For another geometric sequence $${e_n} = – 6,{\text{ }} – 3,{\text{ }} – \frac{3}{2},{\text{ }} – \frac{3}{4},{\text{ }} \ldots$$

find the exact value of the tenth term. Give your answer as a fraction.

b(ii).

## Markscheme

(i)     $${d_n}\;\;\;\;\;$$OR$$\;\;\;1,{\text{ }}0.95,{\text{ }}0.90,{\text{ }}0.85,{\text{ }} \ldots$$     (A1)     (C1)

(ii)     $${b_n}\;\;\;$$OR$$\;\;\;1,{\text{ }}\frac{3}{2},{\text{ }}\frac{9}{4},{\text{ }}\frac{{27}}{8},{\text{ }} \ldots$$     (A1)     (C1)

a.

$$\frac{1}{2}\;\;\;$$OR$$\;\;\;0.5$$     (A1)     (C1)

Note: Accept ‘divide by 2’ for (A1).

b(i).

$$– 6{\left( {\frac{1}{2}} \right)^{10 – 1}}$$     (M1)(A1)(ft)

Notes: Award (M1) for substitution in the GP $${n^{{\text{th}}}}$$ term formula, (A1)(ft) for their correct substitution.

Follow through from their common ratio in part (b)(i).

OR

$$\left( { – 6,{\text{ }} – 3,{\text{ }} – \frac{3}{2},{\text{ }} – \frac{3}{4},} \right) – \frac{3}{8},{\text{ }} – \frac{3}{{16}},{\text{ }} – \frac{3}{{32}},{\text{ }} – \frac{3}{{64}},{\text{ }} – \frac{3}{{128}}$$     (M1)(A1)(ft)

Notes: Award (M1) for terms 5 and 6 correct (using their ratio).

Award (A1)(ft) for terms 7, 8 and 9 correct (using their ratio).

$$– \frac{3}{{256}}\;\;\;\left( { – \frac{6}{{512}}} \right)$$     (A1)(ft)     (C3)

b(ii).

## Question

Only one of the following four sequences is arithmetic and only one of them is geometric.

$${a_n} = 1,{\text{ }}2,{\text{ }}3,{\text{ }}5,{\text{ }} \ldots$$

$${b_n} = 1,{\text{ }}\frac{3}{2},{\text{ }}\frac{9}{4},{\text{ }}\frac{{27}}{8},{\text{ }} \ldots$$

$${c_n} = 1,{\text{ }}\frac{1}{2},{\text{ }}\frac{1}{3},{\text{ }}\frac{1}{4},{\text{ }} \ldots$$

$${d_n} = 1,{\text{ }}0.95,{\text{ }}0.90,{\text{ }}0.85,{\text{ }} \ldots$$

State which sequence is

(i)     arithmetic;

(ii)     geometric.

a.

For another geometric sequence $${e_n} = – 6,{\text{ }} – 3,{\text{ }} – \frac{3}{2},{\text{ }} – \frac{3}{4},{\text{ }} \ldots$$

write down the common ratio;

b(i).

For another geometric sequence $${e_n} = – 6,{\text{ }} – 3,{\text{ }} – \frac{3}{2},{\text{ }} – \frac{3}{4},{\text{ }} \ldots$$

find the exact value of the tenth term. Give your answer as a fraction.

b(ii).

## Markscheme

(i)     $${d_n}\;\;\;\;\;$$OR$$\;\;\;1,{\text{ }}0.95,{\text{ }}0.90,{\text{ }}0.85,{\text{ }} \ldots$$     (A1)     (C1)

(ii)     $${b_n}\;\;\;$$OR$$\;\;\;1,{\text{ }}\frac{3}{2},{\text{ }}\frac{9}{4},{\text{ }}\frac{{27}}{8},{\text{ }} \ldots$$     (A1)     (C1)

a.

$$\frac{1}{2}\;\;\;$$OR$$\;\;\;0.5$$     (A1)     (C1)

Note: Accept ‘divide by 2’ for (A1).

b(i).

$$– 6{\left( {\frac{1}{2}} \right)^{10 – 1}}$$     (M1)(A1)(ft)

Notes: Award (M1) for substitution in the GP $${n^{{\text{th}}}}$$ term formula, (A1)(ft) for their correct substitution.

Follow through from their common ratio in part (b)(i).

OR

$$\left( { – 6,{\text{ }} – 3,{\text{ }} – \frac{3}{2},{\text{ }} – \frac{3}{4},} \right) – \frac{3}{8},{\text{ }} – \frac{3}{{16}},{\text{ }} – \frac{3}{{32}},{\text{ }} – \frac{3}{{64}},{\text{ }} – \frac{3}{{128}}$$     (M1)(A1)(ft)

Notes: Award (M1) for terms 5 and 6 correct (using their ratio).

Award (A1)(ft) for terms 7, 8 and 9 correct (using their ratio).

$$– \frac{3}{{256}}\;\;\;\left( { – \frac{6}{{512}}} \right)$$     (A1)(ft)     (C3)

b(ii).

## Question

The first three terms of a geometric sequence are $${u_1} = 486,{\text{ }}{u_2} = 162,{\text{ }}{u_3} = 54$$.

Find the value of $$r$$, the common ratio of the sequence.

a.

Find the value of $$n$$ for which $${u_n} = 2$$.

b.

Find the sum of the first 30 terms of the sequence.

c.

## Markscheme

$$\frac{{162}}{{486}}$$$$\,\,\,$$OR$$\,\,\,$$$$\frac{{54}}{{162}}$$     (M1)

Note:     Award (M1) for dividing any $${u_{n + 1}}$$ by $${u_n}$$.

$$= \frac{1}{3}{\text{ }}(0.333,{\text{ }}0.333333 \ldots )$$     (A1)     (C2)[2 marks]

a.

$$486{\left( {\frac{1}{3}} \right)^{n – 1}} = 2$$     (M1)

Note:     Award (M1) for their correct substitution into geometric sequence formula.

$$n = 6$$     (A1)(ft)     (C2)

Note:     Follow through from part (a).

Award (A1)(A0) for $${u_6} = 2$$ or $${u_6}$$ with or without working.[2 marks]

b.

$${S_{30}} = \frac{{486\left( {1 – {{\frac{1}{3}}^{30}}} \right)}}{{1 – \frac{1}{3}}}$$     (M1)

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

$$= 729$$     (A1)(ft)     (C2)[2 marks]

c.