# IBDP Maths AI: Topic : SL 1.2: Arithmetic sequences and series: IB style Questions SL Paper 1

## Question

The fifth term of an arithmetic sequence is 20 and the twelfth term is 41.

(i) Find the common difference.

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

a.

Calculate the eighty-fourth term.[1]

b.

Calculate the sum of the first 200 terms.[2]

c.

## Markscheme

(i) $$u_5 = u_1 + 4d = 20$$

$$u_{12} = u_1 + 11d = 41$$     (M1)

(M1) for both equations correct (or (M1) for $$20 + 7d = 41$$)

$$7d = 21$$

$$d = 3$$     (A1)     (C2)

(ii) $$u_1 + 12 = 20$$

$$u_1 = 8$$     (A1)(ft)     (C1) [3 marks]

a.

$$u_{84} = 8 + (84 – 1)3$$

$$= 257$$     (A1)(ft)     (C1)[1 mark]

b.

$$S_{200} = 100(16 + 199 \times 3)$$     (M1)

$$= 61300$$     (A1)(ft)     (C2)[2 marks]

c.

## 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

$${u_{96}} = {u_1} + 95d$$     (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

Given the arithmetic sequence: $${u_1} = 124{\text{, }}{u_2} = 117{\text{, }}{u_3} = 110{\text{, }}{u_4} = 103, \ldots$$

Write down the common difference of the sequence.[1]

a.

Calculate the sum of the first $$50$$ terms of the sequence.[2]

b.

$${u_k}$$ is the first term in the sequence which is negative.

Find the value of $$k$$.[3]

c.

## Markscheme

$$d = – 7$$     (A1)     (C1)[1 mark]

a.

$${S_{50}} = \frac{{50}}{2}(2(124) + 49( – 7))$$     (M1)

Note: (M1) for correct substitution.

$$= – 2375$$     (A1)(ft)     (C2)[2 marks]

b.

$$124 – 7(k – 1) < 0$$     (M1)

$$k > 18.7$$ or $$18.7$$ seen     (A1)(ft)

$$k = 19$$     (A1)(ft)     (C3)

Note: (M1) for correct inequality or equation seen or for list of values seen or for use of trial and error.[3 marks]

c.

## Question

A tree begins losing its leaves in October. The number of leaves that the tree loses each day increases by the same number on each successive day.

Calculate the number of leaves that the tree loses on the 21st October.[3]

a.

Find the total number of leaves that the tree loses in the 31 days of the month of October.[3]

b.

## Markscheme

$$u_{21} = 24 + (21 – 1)(16)$$     (M1)(A1)

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

$$u_{21} = 344$$     (A1)     (C3)[3 marks]

a.

$${S_{31}} = \frac{{31}}{2}\left[ {2(24) + (31 – 1)(16)} \right]$$     (M1)(A1)(ft)

Note: Award (M1) for correct substituted formula, (A1)(ft) for correct substitutions. (ft) from their value for d.

$$S_{31} = 8184$$     (A1)(ft)     (C3)[3 marks]

b.

## Question

Consider the following sequence:

57, 55, 53 …, 5, 3

Find the number of terms of the sequence.[3]

a.

Find the sum of the sequence.[3]

b.

## Markscheme

3 = 57 + (n − 1) × (−2)

OR

57 = 3 + (n − 1) × (2)     (A1)(M1)

Note: Award (A1) for 3 or 57 seen as un, (M1) for correctly substituted formula or list of values seen.

n = 28     (A1)     (C3)[3 marks]

a.

$${{\text{S}}_{28}} = \frac{{28}}{2}(57 + 3)$$

OR

$${{\rm{S}}_{28}} = \frac{{28}}{2}(2(57) + (28 – 1) \times – 2)$$

OR

$${{\rm{S}}_{28}} = \frac{{28}}{2}(2(3) + (28 – 1) \times 2)$$     (M1)(A1)(ft)

Note: (A1)(ft) for 28 seen.

Award (M1) for correctly substituted formula or list of values seen.

$${{\rm{S}}_{28}} = 840$$     (A1)(ft)     (C3)[3 marks]

b.

## Question

The first term of an arithmetic sequence is 7 and the sixth term is 22.

Find the common difference.[2]

a.

Find the twelfth term.[2]

b.

Find the sum of the first 100 terms.[2]

c.

## Markscheme

$$7 + 5d = 22$$     (M1)

Note: Award (M1) for correct substitution in the AP formula. Accept list of numbers as solution.

$$d = 3$$     (A1)     (C2)[2 marks]

a.

$$u_{12} = 7 + 11 \times 3$$     (M1)

$$= 40$$     (A1)(ft)     (C2)

Note: Accept list of numbers.[2 marks]

b.

$$S_{100} = \frac{{100}}{{2}} (2 \times 7 + 99 \times 3)$$     (M1)

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

= 15550     (A1)(ft)     (C2)

Note: Accept 15600[2 marks]

c.

## Question

A concert choir is arranged, per row, according to an arithmetic sequence. There are 20 singers in the fourth row and 32 singers in the eighth row.

Find the common difference of this arithmetic sequence.[3]

a.

There are 10 rows in the choir and 11 singers in the first row.

Find the total number of singers in the choir.[3]

b.

## Markscheme

20 = u1 + 3d     (A1)

32 = u1 + 7d     (A1)

Note: Award (A1) for each equation, (A1) for correct answer.

OR

$$d = \frac{{32 – 20}}{4}$$     (A1)(A1)

Note: Award (A1) for numerator, (A1) for denominator.

d = 3     (A1)     (C3)[3 marks]

a.

$$\frac{{10}}{2}(2 \times 11 + 9 \times 3)$$ or $$\frac{{10}}{2}(11 + 38)$$     (M1)(A1)(ft)

Note: Award (M1) for correct substituted formula, (A1) for correct substitution, follow through from their answer to part (a).

OR

11 + 14 + … + 38     (M1)(A1)(ft)

Note: Award (M1) for attempt at the sum of a list, (A1)(ft) for all correct numbers, follow through from their answer to part (a).

= 245     (A1)(ft)     (C3)[3 marks]

b.

## Question

The first term of an arithmetic sequence is 3 and the sum of the first two terms is 11.

Write down the second term of this sequence.[1]

a.

Write down the common difference of this sequence.[1]

b.

Write down the fourth term of this sequence.[1]

c.

The nth term is the first term in this sequence which is greater than 1000. Find the value of n.[3]

d.

## Markscheme

8     (A1)     (C1)[1 mark]

a.

5     (A1)(ft)     (C1)[1 mark]

b.

18     (A1)(ft)     (C1)[1 mark]

c.

3 + 5 × (n – 1) > 1000     (M1)

Note: Allow equality sign and equal to 1001

n > 200.4     (A1)

Note: Accept n = 200.4 or 5n =1002

OR

(M1) for attempt at listing, (A1) for 998 and 1003 seen.     (M1)(A1)

n = 201     (A1)(ft)     (C3)

d.

## Question

A teacher earns an annual salary of $$45 000$$ USD for the first year of her employment. Her annual salary increases by $$1750$$ USD each year.

Calculate the annual salary for the fifth year of her employment.[3]

a.

She remains in this employment for 10 years. Calculate the total salary she earns in this employment during these 10 years.[3]

b.

## Markscheme

$$45000 + (5 – 1)1750$$     (M1)(A1)

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

$$= 52000$$ USD     (A1)     (C3)

Notes: If a list is used, award (M1) for recognizing AP, award (A1) for seeing 52000 in their list, (A1) for final answer.[3 marks]

a.

$$\frac{{10}}{2}(2(45000) + (10 – 1)(1750))$$     (M1)(A1)

Notes: Award (M1) for substituted AP formula, (A1)(ft) for correct substitutions. Follow through from their common difference used in part (a).

$$= 528750$$ USD     (A1)(ft)     (C3)

Notes: Accept $$529 000$$. If a list is used, award (M1) for recognizing sum of AP, (A1) for seeing $$60 750$$ included in the sum or $$528 750$$ in a cumulative list.[3 marks]

b.

## Question

Consider the arithmetic sequence
${\text{326, 321, 316, 311, }} \ldots {\text{, 191.}}$

Find the value of the common difference of this sequence.[2]

a.

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

b.

Find the number of terms in this sequence.[2]

c.

## Markscheme

$$d = 321 – 326$$ (or equivalent)

$$= – 5$$     (A1)(A1)     (C2)

Note: Award (A1) for negative sign. (A1) for 5.[2 marks]

a.

$${S_{10}} = \frac{{10}}{2}(2(326) + 9( – 5))$$     (M1)

Notes: Award (M1) for correctly substituted formula. Follow through from part (a).

OR

$${u_{10}} = 281$$

$${S_{10}} = \frac{{10}}{2}(326 + 281)$$     (M1)

Note: Award (M1) for correctly substituted formula, not for finding 281.

OR

If a list is seen award (M1) for the correct list of 10 terms consistent with their $$d$$.     (M1)

$$= 3035$$     (A1)(ft)     (C2)

Note: If $$d = 5$$ final answer is 3485. Follow through from part (a). No follow through if list used.[2 marks]

b.

$$191 = 326 + (n – 1)( – 5)$$     (or equivalent)     (M1)

Notes: Award (M1) for correctly substituted formula. Follow through from part (a).

OR

If a list is seen award (M1) for the complete and correct list of terms or complete list of terms consistent with their $$d$$.     (M1)

$$n = 28$$     (A1)(ft)     (C2)

Note: $$n$$ must be a positive integer. Follow through from part (a). No follow through if list used.[2 marks]

c.

## Question

The tenth term of an arithmetic sequence is 32 and the common difference is –6.

Find the first term of the sequence.[2]

a.

Find the 21st term of the sequence.[2]

b.

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

c.

## Markscheme

32 = u1 + (10 − 1) × (−6)     (M1)

Notes: Award (M1) for correct substitution in correct formula. Accept correct alternative methods.

u1 = 86     (A1)     (C2)[2 marks]

a.

u21 = 86 + (21 − 1) × (−6)     (M1)

u21 = −34     (A1)(ft)

Notes: Award (M1) for correct substitution in correct formula. Accept correct alternative methods. Award (M1) for a list of at least 5 correct terms seen. Follow through from their answer to part (a).

OR

u21 = 32 + 11 × (−6)     (M1)

u21 = −34     (A1)     (C2)[2 marks]

b.

$${S_{30}} = \frac{{30}}{2}(2 \times 86 + (30 – 1) \times ( – 6))$$     (M1)

Notes: Award (M1) for their correct substitution in correct formula. Accept correct alternative methods. For a list award (M1) for the correct addition of at least 10 terms.

$${S_{30}} = -30$$     (A1)(ft)     (C2)

c.

## Question

The first term of an arithmetic sequence is 3 and the seventh term is 33.

Calculate the common difference;[2]

a.

Calculate the 95th term of the sequence;[2]

b.

Calculate the sum of the first 250 terms.[2]

c.

## Markscheme

33 = 3 + d(6)     (M1)

Note: Award (M1) for correctly substituted formula or a correct numerical expression to find the common difference.

(d =)5     (A1)     (C2)[2 marks]

a.

u95 = 3 + 94(5)     (M1)

Note: Award (M1) for their correctly substituted formula.

= 473     (A1)(ft)     (C2)

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

b.

S250 = 125[2(3) + 249(5)]     (M1)

Note: Award (M1) for correctly substituted formula.

S250 = 156375     (A1)(ft)     (C2)

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

c.

## Question

In an arithmetic sequence, the fifth term, u5, is greater than the first term, u1. The difference between these terms is 36.

Find the common difference, d.[2]

a.

The tenth term of the sequence is double the seventh term.

(i) Write down an equation in u1 and d to show this information.

(ii) Find u1.[4]

b.

## Markscheme

$${u_1} + 4d – {u_1} = 36$$     (M1)

Note: Accept equivalent forms including the use of a instead of $${u_1}$$.

$$(d =) 9$$     (A1)     (C2)

a.

(i) $${u_{10}} = 2{u_7}$$     (M1)

Note: Award (M1) for correct use of 2 (may be implied).

$${u_1} + 9d = 2[{u_1} + 6d]$$     (A1)

Notes: Accept equivalent forms. Award (M1)(A0) for $$a + 9d = 2[a + 6d]$$.

(ii) $${u_1} + 81 = 2{u_1} + 108$$     (M1)

$$({u_1} =) – 27$$     (A1)(ft)     (C4)

Notes: Follow through from their d found in part (a) and equation in (b)(i). Do not penalize further use of a instead of $${u_1}$$.

b.

## Question

The first term, $${u_1}$$, of an arithmetic sequence is $$145$$. The fifth term, $${u_5}$$, of the sequence is $$113$$.

Find the common difference of the sequence.[2]

a.

The $${n^{{\text{th}}}}$$ term, $${u_n}$$, of the sequence is $$–7$$.

Find the value of $$n$$.[2]

b.

The $${n^{{\text{th}}}}$$ term, $${u_n}$$, of the sequence is $$–7$$.

Find $${S_{20}}$$, the sum of the first twenty terms of the sequence.[2]

c.

## Markscheme

$$145 + (5 – 1)d = 113$$     (M1)

Note: Award (M1) for correctly substituted AP formula.

OR

$$\frac{{113 – 145}}{4}$$     (M1)

$$= – 8$$     (A1)     (C2)[2 marks]

a.

$$145 + (n – 1) \times – 8 = – 7$$     (M1)

Note: Award (M1) for their correctly substituted AP formula.

If a list is used award (M1) for their correct values down to $$−7$$.

$$n = 20$$     (A1)(ft)     (C2)

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

b.

$${S_{20}} = \frac{{20}}{2}\left( {2 \times 145 + (20 – 1) \times – 8} \right)$$     (M1)

Note: Award (M1) for their correctly substituted sum of an AP formula.

If a list is used award (M1) for their correct terms up to $$1380$$

$$= 1380$$     (A1)(ft)

Note: Follow through from their part (a).

OR

$${S_{20}} = \frac{{20}}{2}\left( {145 + ( – 7)} \right)$$     (M1)

Note: Award (M1) for correctly substituted sum of an AP formula.

$$= 1380$$     (A1)     (C2)

Note: If candidates have listed the terms correctly and given the common difference as $$8$$, award (M1)(A0) for part (a), (M1)(A0) for an answer of $$−18$$ or $$18$$ for part (b) and (M1)(A1)(ft) for an answer of $$4420$$ in part (c) with working seen.[2 marks]

c.

## Question

The number of apartments in a housing development has been increasing by a constant amount every year.

At the end of the first year the number of apartments was 150, and at the end of the sixth year the number of apartments was 600.

The number of apartments, $$y$$, can be determined by the equation $$y = mt + n$$, where $$t$$ is the time, in years.

Find the value of $$m$$.[2]

a.

State what $$m$$ represents in this context.[1]

b.

Find the value of $$n$$.

[2]
c.

State what $$n$$ represents in this context.[1]

d.

## Markscheme

$$\frac{{600 – 150}}{{6 – 1}}$$     (M1)

OR

$$600 = 150 + (6 – 1)m$$     (M1)

Note: Award (M1) for correct substitution into gradient formula or arithmetic sequence formula.

$$= 90$$     (A1)     (C2)

a.

the annual rate of growth of the number of apartments     (A1)     (C1)

Note: Do not accept common difference.

b.

$$150 = 90 \times (1) + n$$     (M1)

Note: Award (M1) for correct substitution of their gradient and one of the given points into the equation of a straight line.

$$n = 60$$     (A1)(ft)     (C2)

Note: Follow through from part (a).

c.

the initial number of apartments     (A1)     (C1)

Note: Do not accept “first number in the sequence”.

d.

## 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.[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)     $${u_1} + d = 30,{\text{ }}{u_1} + 4d = 90,{\text{ }}3d = 90 – 30\;\;\;$$(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.[2]

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;[1]

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.[3]

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

A comet orbits the Sun and is seen from Earth every 37 years. The comet was first seen from Earth in the year 1064.

Find the year in which the comet was seen from Earth for the fifth time.[3]

a.

Determine how many times the comet has been seen from Earth up to the year 2014.[3]

b.

## Markscheme

$$1064 + (5 – 1) \times 37$$    (M1)(A1)

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

$$= 1212$$    (A1)     (C3)[3 marks]

a.

$$2014 > 1064 + (n – 1) \times 37$$    (M1)

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

Accept an equation.

$$(n < ){\text{ }}26.6756 \ldots$$    (A1)

26 (times)     (A1)     (C3)

Note:     Award the final (A1) for the correct rounding down of their unrounded answer.

OR

$$2014 > 1064 + 37t$$    (M1)

Note:     Award (M1) for a correct substitution into a linear model (where $$t = n – 1$$).

Accept an equation or weak inequality.

Accept $$\frac{{2014 – 1064}}{{37}}$$ for     (M1).

$$(t < ){\text{ }}25.6756 \ldots$$    (A1)

26 (times)     (A1)     (C3)

Note:     Award the final (A1) for adding 1 to the correct rounding down of their unrounded answer.[3 marks]

b.