# 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.[2]

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.[2]

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.[2]

c.

## Markscheme

$${u_{96}} = {u_1} + 95d$$Â Â Â Â  (M1)

$$= 0 + 95 \times 12$$

$$= 1140$$Â Â Â Â  (A1)Â Â Â Â  (C2)

[2 marks]

a.

$$6{r^5} = 16d$$ Â Â Â  (A1)

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

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

b.

$${r^5} = 32$$Â Â Â Â  (A1)(ft)

Note: (ft) from their (b).

$$r = 2$$Â Â Â Â  (A1)(ft)Â Â Â Â  (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.[2]

a.

The fees continue to increase in the same ratio.

Find the fees paid for 2006.[2]

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.[2]

c.

## Markscheme

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

Note: Award (M1) for dividing correct terms.

r = 1.04Â  Â Â  (A1)Â Â Â Â  (C2)

Notes: In (b) and (c) (ft) from candidateâ€™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)6Â Â Â Â  (M1)

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

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

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

Give full credit for solution by lists.

(FP) Â  Â  Total = 73713.81 USD (USD not required) Â  Â  (A1)(ft) Â  Â  (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.[1]

a.

Find the population of big cats at the beginning of 2010.[2]

b.

Find the number of years, from the beginning of 2004, it will take the population of big cats to exceed $$50\,000$$.[3]

c.

## Markscheme

$$10\,000 \times 1.05$$

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

a.

$$10\,000 \times {1.05^6}$$ Â  Â  (M1)

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

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

b.

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

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

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

c.

## Question

Consider the geometric sequence 16, 8, a, 2, b, â€¦

Write down the common ratio.[1]

a.

Write down the value ofÂ a.[1]

b.i.

Write down the value of b.[1]

b.ii.

The sum of the first n terms is 31.9375. Find the value of n.[3]

c.

## Markscheme

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

a.

4Â Â Â Â  (A1)[1 mark]

b.i.

1 Â  Â  (A1)Â Â Â Â  (C2)[1 mark]

b.ii.

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

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

OR

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

indication of point where y = 31.9375Â Â Â Â  (M1)

OR

16 + 8 + 4 + 2 + 1 + 0.5 + 0.25 + 0.125 + 0.0625 = 31.9375Â Â Â Â  (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 = 9Â Â Â Â  (A1)(ft) Â  Â  (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.[1]

a.

Find $${u_1}$$ .[2]

b.

The sum of the first $$k$$ terms in the sequence is $$118 096$$ . Find the value of $$k$$ .[3]

c.

## Markscheme

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

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

a.

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

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

$${u_1} = 78732$$ Â  Â  (A1)(ft) Â  Â  (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.

$$118096 = \frac{{78732\left( {1 – {{\left( {\frac{1}{3}} \right)}^k}} \right)}}{{\left( {1 – \frac{1}{3}} \right)}}$$Â Â Â Â  (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 $$y = 78732\frac{{\left( {1 – {{\left( {\frac{1}{3}} \right)}^k}} \right)}}{{\left( {1 – \frac{1}{3}} \right)}}$$Â Â Â Â  (M1)
Indication of point where $$y = 118 096$$ Â  Â  (M1)

OR

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

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

$$k =10$$ Â  Â  (A1)(ft)Â Â Â Â  (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.[3]

a.

Calculate the least number of terms required for the sum of the sequence to be greater than 682.6[3]

b.

## Markscheme

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

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

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

a.

$$\frac{{512\left( {1 – {{\left( {\frac{1}{4}} \right)}^n}} \right)}}{{1 – \left( {\frac{1}{4}} \right)}} > 682.6$$ Â Â Â  (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 donâ€™t have to be correct.

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

$$n = 7$$ Â Â Â  (A1)(ft) Â  Â  (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.[1]

a.

Calculate the total distance Javier runs in the first seven days of training.[2]

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.[3]

c.

## Markscheme

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

a.

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

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

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

b.

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

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

n = 21.371… Â  Â  (A1)(ft)

n = 22 Â  Â  (A1)(ft) Â  Â  (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;[2]

a.

Calculate the eleventh term of the sequence;[2]

b.

Calculate the sum of the first 23 terms of the sequence.[2]

c.

## Markscheme

2r2 = 2.205Â Â Â Â  (M1)

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

r = 1.05 Â  Â  (A1) Â  Â  (C2)[2 marks]

a.

2(1.05)10Â Â Â Â  (M1)

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

= 3.26 Â  Â  (3.25778â€¦) Â  Â  (A1)(ft)Â Â Â Â  (C2)

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

b.

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

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

= 82.9 Â  Â  (82.8609â€¦)Â Â Â Â  (A1)(ft) Â  Â  (C2)

Notes: Accept an answer of 3.97221…if r = âˆ’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.[2]

a.

Find u1, the first term of the sequence.[2]

b.

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

c.

## Markscheme

$$\frac{{101.25}}{{135}}$$ Â Â Â  (M1)

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

a.

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

OR

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

OR

(by list)

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

Â

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

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

b.

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

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

Â Â Â  Accept a list of all their ten geometric terms.

= 1210 (1207.918…) Â  Â  (A1)(ft) Â  Â  (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.[3]

a.

Find the total number of matches played in the tournament.[3]

b.

## Markscheme

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

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

Â Â  Â  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.

$$16$$ Â  Â Â (A1) Â  Â  (C3)[3 marks]

a.

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

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

Â Â  Â  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.

$$511$$ Â  Â Â (A1) Â  Â  (C3)[3 marks]

b.

## Question

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

Calculate

(i) Â  Â  the common difference of the sequence;

(ii) Â  Â  the first term of the sequence.[3]

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.[3]

b.

## Markscheme

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

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

$$(d = ){\text{ }}20$$ Â  Â  (A1)

(ii) Â  Â  $$({u_1} = ){\text{ }}10$$ Â  Â  (A1)(ft) Â  Â  (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$$\;\;\;({u_7} = ){\text{ 10}} \times {3^6}$$Â Â  Â  (M1)(A1)(ft)

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

OR

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

Note:Â 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).

$$= 7290$$ Â  Â  (A1)(ft) Â  Â  (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.

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

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

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

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

State which sequence is

(i) Â  Â  arithmetic;

(ii) Â  Â  geometric.[2]

a.

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

write down the common ratio;[1]

b(i).

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

find the exact value of the tenth term. Give your answer as a fraction.[3]

b(ii).

## Markscheme

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

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

a.

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

Note: Accept â€˜divide by 2â€™ for (A1).

b(i).

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

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

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

OR

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

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

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

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

b(ii).

## Question

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

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

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

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

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

State which sequence is

(i) Â  Â  arithmetic;

(ii) Â  Â  geometric.[2]

a.

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

write down the common ratio;[1]

b(i).

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

find the exact value of the tenth term. Give your answer as a fraction.[3]

b(ii).

## Markscheme

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

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

a.

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

Note: Accept â€˜divide by 2â€™ for (A1).

b(i).

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

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

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

OR

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

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

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

$$– \frac{3}{{256}}\;\;\;\left( { – \frac{6}{{512}}} \right)$$ Â  Â  (A1)(ft) Â  Â  (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.[2]

a.

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

b.

Find the sum of the first 30 terms of the sequence.[2]

c.

## Markscheme

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

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

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

a.

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

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

$$n = 6$$ Â  Â  (A1)(ft) Â  Â  (C2)

Note: Â  Â  Follow through from part (a).

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

b.

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

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

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

c.