1

1. 17 + 27 + 37 + ... + 417 17 + (n – 1)10 = 417 (M1) 10(n – 1) = 400 n = 41 (A1)

S41 = 241

(2(17) + 40(10)) (M1)

= 41(17 + 200) = 8897 (A1)

OR

S41 = 241

(17 + 417) (M1)

= 241

(434)

= 8897 (A1) (C4) [4]

2. Required term is !!"

#$$%

&

58

(3x)5(–2)3 (A1)(A1)(A1)

Therefore the coefficient of x5 is 56 × 243 × –8 = –108864 (A1) (C4)

[4]

3. S5 = 25

{2 + 32} (M1)(A1)(A1)

S5 = 85 (A1) OR a = 2, a + 4d = 32 (M1) ⇒ 4d = 30 d = 7.5 (A1)

S5 = 25

(4 + 4(7.5)) (M1)

= 25

(4 + 30)

S5 = 85 (A1) (C4) [4]

2

4. (5a + b)7 = ...+ !!"

#$$%

&

47

(5a)3(b)4 + ... (M1)

= 43214567

×××

××× × 53 × (a3b4) = 35 × 53 × a3b4 (M1)(A1)

So the coefficient is 4375 (A1) (C4) [4]

5. 43x–1 = 1.5625 × 10–2 (3x – 1)log10 4 = log10 1.5625 – 2 (M1)

⇒ 3x – 1 = 4log

25625.1log

10

10 − (A1)

⇒ 3x – 1 = –3 (A1)

⇒ x = –32

(A1) (C4)

[4]

6. a = 5 a + 3d = 40 (may be implied) (M1)

d = 335

(A1)

T2 = 5 + 335

(A1)

= 1632

or 350

or 16.7 (3 sf) (A1) (C4)

[4]

7. (a) log2 5 = 2log5log

a

a (M1)

= xy

(A1) (C2)

(b) loga 20 = loga 4 + loga 5 or loga 2 + loga 10 (M1) = 2 loga 2 + loga 5 = 2x + y (A1) (C2)

[4]

8. S = !"

#$%

&−−

=−

321

32

11

ru

(M1)(A1)

3

= 53

32× (A1)

= 52

(A1) (C4)

[4]

9. (a + b)12

Coefficient of a5b7 is !!"

#$$%

&=!!

"

#$$%

&

712

512

(M1)(A1)

= 792 (A2) (C4) [4]

10. (a) Plan A: 1000, 1080, 1160... Plan B: 1000, 1000(1.06), 1000(1.06)2… 2nd month: $1060, 3rd month: $1123.60 (A1)(A1) 2

(b) For Plan A, T12 = a + 11d = 1000 + 11(80) (M1) = $1880 (A1)

For Plan B, T12 = 1000(1.06)11 (M1) = $1898 (to the nearest dollar) (A1) 4

(c) (i) For Plan A, S12 = 212

[2000 + 11(80)] (M1)

= 6(2880) = $17280 (to the nearest dollar) (A1)

(ii) For Plan B, S12 = 106.1

)106.1(1000 12

−

− (M1)

= $16870 (to the nearest dollar) (A1) 4 [10]

11. (a) $1000 × 1.07510 = $2061 (nearest dollar) (A1) (C1)

4

(b) 1000(1.07510 + 1.0759 + ... + 1.075) (M1)

= 1075.1

)1075.1)(075.1(1000 10

−

− (M1)

= $15208 (nearest dollar) (A1) (C3) [4]

12. log10

2

3 !!"

#$$%

&

QRP

= 2log10 !!"

#$$%

&3QRP

(M1)

2log10 !!"

#$$%

&3QRP

= 2(log10P – log10(QR3)) (M1)

= 2(1og10P – log10Q – 3log10R) (M1) = 2(x – y – 3z) = 2x – 2y – 6z or 2(x – y – 3z) (A1)

[4]

13. (a) a1 = 1000, an = 1000 + (n – 1)250 = 10000 (M1)

n = 250100000010 −

+ 1 = 37.

She runs 10 km on the 37th day. (A1)

(b) S37 = 237

(1000 + 10000) (M1)

She has run a total of 203.5 km (A1) [4]

14. The constant term will be the term independent of the variable x. (R1) 9

2

3

26

289

9

22...2

39

...292!"

#$%

& −++!

"

#$%

& −!!"

#$$%

&++!

"

#$%

& −+=!

"

#$%

&−

xxx

xxx

xx (M1)

!"

#$%

& −=!

"

#$%

& −!!"

#$$%

&6

63

26 8842

39

xx

xx (A1)

= –672 (A1) [4]

5

15. (3x + 2y)4 = (3x)4 + !!"

#$$%

&

14

(3x)2(2y) + !!"

#$$%

&

24

(3x)2(2y)2 + !!"

#$$%

&

34

(3x)(2y)3 +(2y)4 (A1)

= 81x4 + 216x3y + 216x2y2 + 96xy3 + 16y4 (A1)(A1)(A1) (C4) [4]

16. (a) u1 = 7, d = 2.5 (M1) u41 = u1 + (n – 1)d = 7 + (41 – 1)2.5 = 107 (A1) (C2)

(b) S101 = 2n

[2u1 + (n – 1)d]

= 2101

[2(7) + (101 – 1)2.5] (M1)

= 2

)264(101

= 13332 (A1) (C2) [4]

17. METHOD 1

log9 81 + log9 !"

#$%

&91

+ log9 3 = 2 – 1 + 21

(M1)

⇒ 23

= log9 x (A1)

⇒ x = 23

9 (M1) ⇒ x = 27 (A1) (C4)

METHOD 2

log 81 + log9 !"

#$%

&91

+ log9 3 = log9 !"

#$%

&'(

)*+

, 39181 (M2)

= log9 27 (A1) ⇒ x = 27 (A1) (C4)

[4]

6

18. (a) (1 + 1)4 = 24 = 1 + )1(34

)1(24

)1(14 32

!!"

#$$%

&+!!

"

#$$%

&+!!

"

#$$%

& + 14 (M1)

⇒ !!"

#$$%

&+!!

"

#$$%

&+!!

"

#$$%

&

34

24

14

= 16 – 2

= 14 (A1) (C2)

(b) (1 + 1)9 = 1 + !!"

#$$%

&++!!

"

#$$%

&+!!

"

#$$%

&+!!

"

#$$%

&

89

...39

29

19

+ 1 (M1)

⇒ !!"

#$$%

&++!!

"

#$$%

&+!!

"

#$$%

&+!!

"

#$$%

&

89

...39

29

19

= 29 – 2

= 510 (A1) (C2) [4]

19. (a) r = 23

160240

240360

== = 1.5 (A1) 1

(b) 2002 is the 13th year. (M1) u13 = 160(1.5)13–1 (M1) = 20759 (Accept 20760 or 20800.) (A1) 3

(c) 5000 = 160(1.5)n–1

1605000

= (1.5)n–1 (M1)

log !"

#$%

&1605000

= (n – 1)log1.5 (M1)

n – 1 = 5.1log

1605000log !

"

#$%

&

= 8.49 (A1)

⇒ n = 9.49 ⇒ 10th year ⇒ 1999 (A1)

OR

Using a gdc with u1 = 160, uk+1 = 23 uk, u9 = 4100, u10 = 6150 (M2)

1999 (G2) 4

7

(d) S13 = 160 !"

#$%

&

−

−

15.115.1 13

(M1)

= 61958 (Accept 61960 or 62000.) (A1) 2

(e) Nearly everyone would have bought a portable telephone so there would be fewer people left wanting to buy one. (R1)

OR

Sales would saturate. (R1) 1 [11]

20. (a) u4 = ul + 3d or 16 = –2 +3d (M1)

d = ( )32––16

(M1)

= 6 (A1) (C3)

(b) un = ul + (n – 1)6 or 11998 = –2 + (n – l)6 (M1)

n = 16

211998+

+ (A1)

= 2001 (A1) (C3) [6]

21. (a) 10 (A2) (C2)

(b) (3x2)3 61– !"

#$%

&x

[for correct exponents] (M1)(A1)

!!"

#$$%

&

69

33 x6 !"

#$%

& × 663

6

13 84or

1x

xx

(A1)

constant = 2268 (A1) (C4) [6]

22. log27 (x(x – 0.4)) = l (M1)(A1) x2 – 0.4x = 27 (M1) x = 5.4 or x = –5 (G2) x = 5.4 (A1) (C6)

Note: Award (C5) for giving both roots. [6]

23.

8

Statement (a) Is the statement true for all real numbers x? (Yes/No)

(b) If not true, example

A No x = –l (log10 0.1 = –1) (a) (A3) (C3)

B No x = 0 (cos 0 = 1) (b) (A3) (C3)

C Yes N/A

Notes: (a) Award (A1) for each correct answer. (b) Award (A) marks for statements A and B only if NO in column (a). Award (A2) for a correct counter example to statement A, (A1) for a correct counter example to statement B (ignore other incorrect examples). Special Case for statement C: Award (A1) if candidates write NO, and give a valid reason (eg

arctan 1 = 4π5

).

[6]

24. (a) Ashley AP 12 + 14 + 16 + ... to 15 terms (M1)

S15 = 215

[2(12) + 14(2)] (M1)

= 15 × 26 = 390 hours (A1) 3

(b) Billie GP 12, 12(1.1), 12(1.1)2… (M1)

(i) In week 3, 12(1.1)2 (A1) = 14.52 hours (AG)

(ii) S15 = ( )

1–1.1]1–1.1[12 15

(M1)

= 381 hours (3 sf) (A1) 4

9

(c) 12 (1.1)n–1 > 50 (M1)

(1.1)n–1 > 1250

(A1)

(n – 1) ln 1.1 > ln1250

n – 1 > 1.1ln1250ln

(A1)

n – 1 > 14.97 n > 15.97 ⇒ Week 16 (A1)

OR 12(1.1)n–1 > 50 (M1) By trial and error 12(1.1)14 = 45.6, 12(1.1)15 = 50.1 (A1) ⇒ n – l = 15 (A1) ⇒ n = 16 (Week 16) (A1) 4

[11]

25. Term involving x3 is !!"

#$$%

&

35

(2)2(–x)3 (A1)(A1)(A1)

!!"

#$$%

&

35

= 10 (A1)

Therefore, term = –40x3 (A1) ⇒ The coefficient is –40 (A1) (C6)

[6]

26. (a) (i) PQ = 22 AQAP + (M1)

= 22 22 + = ( )24 = 2 2 cm (A1)(AG)

(ii) Area of PQRS = (2 2 )(2 2 ) = 8 cm2 (A1) 3

(b) (i) Side of third square = ( ) ( )2222 + = √4 = 2 cm

Area of third square = 4 cm2 (A1)

10

(ii) 48

32

816

21

rd

nd

nd

st

== (M1)

⇒ Geometric progression, r = 21

84

168

== (A1) 3

(c) (i) u11 = u1r10 = 1610

21!"

#$%

& = 102416

(M1)

= 641

( = 0.015625 = 0.0156, 3 sf) (A1)

(ii) S∞ = r

u–11 =

21–1

16 (M1)

= 32 (A1) 4 [10]

27. Arithmetic sequence d = 3 (may be implied) (M1)(A1) n = 1250 (A2)

S = 2

1250(3 + 3750) !

"

#$%

&×+= 3) 1249 (6

21250 or S (M1)

= 2 345 625 (A1) (C6) [6]

28. Selecting one term (may be implied) (M1)

!"

#$%

&27

52(2x2)5 (A1)(A1)(A1)

= 16800x10 (A1)(A1) (C6) Note: Award C5 for 16800.

[6]

11

29. x f Σf

4 2 2

5 5 7

6 4 11

7 3 14

8 4 18

10 2 20

12 1 21

(a) m = 6 (A2) (C2)

(b) Q1 = 5 (A2) (C2)

(c) Q3 = 8 (A1) IQR = 8 – 5 (M1) = 3 (accept 5 – 8 or [5, 8]) (C2)

[6]

30. (a) log5 x2 = 2 log5 x (M1) = 2y (A1) (C2)

(b) log5x1

= –log5 x (M1)

= –y (A1) (C2)

(c) log25 x = 25log

log

5

5 x (M1)

= y21

(A1) (C2)

[6]

12

31. ... + 6 × 22(ax)2 + 4 × 2(ax)3 + (ax)4 (M1)(M1)(M1) = ...+ 24a2x2 +8a3x3 + a4x4 (A1)(A1)(A1) (C6)

Notes: Award C3 if brackets omitted, leading to 24ax2 + 8ax3+ ax4. Award C4 if correct expression with brackets as in first line of markscheme is given as final answer.

[6]

32. Arithmetic sequence (M1) a = 200 d = 30 (A1)

(a) Distance in final week = 200 + 51 × 30 (M1) = 1730 m (A1) (C3)

(b) Total distance = 252

[2.200 + 51.30] (M1)

= 50180 m (A1) (C3) Note: Penalize once for absence of units ie award A0 the first time units are omitted, A1 the next time.

[6]

33. (a) (i) Area B = 161

, area C = 641

(A1)(A1)

(ii) 41

161641

41

41161

== (Ratio is the same.) (M1)(R1)

(iii) Common ratio = 41

(A1) 5

13

(b) (i) Total area (S2) = 165

161

41

=+ = (= 0.3125) (0.313, 3 sf) (A1)

(ii) Required area = S8 =

411

411

41 8

−

""

#

$

%%

&

'"#

$%&

'−

(M1)

= 0.333328 2(471...) (A1) = 0.333328 (6 sf) (A1) 4

Note: Accept result of adding together eight areas correctly.

(c) Sum to infinity =

411

41

− (A1)

= 31

(A1) 2

[11]

34. METHOD 1

log10 !!"

#$$%

&

zyx2

= log10 x – log10 y2 – log10 z (A1)(A1)(A1)

log10 y2 = 2 log10 y (A1)

log10 z = 21

log z (A1)

log10 !!"

#$$%

&

zyx2

= log10 x – 2log y – 21

log z

= p – 2q – 21 r (A1) (C2)(C2)(C2)

14

METHOD 2

x = 10, y2 = 102p, 210r

z = (A1)(A1)(A1)

log10!!!

"

#

$$$

%

&=!

!"

#$$%

&

22102

1010

10log

rq

p

zyx (A1)

= log10 !"

#$%

&−−=

!!

"

#

$$

%

& −−

2210 2

2 rqprqp

(A2) (C2)(C2)(C2)

[6]

35. (a) (i) Neither

(ii) Geometric series

(iii) Arithmetic series

(iv) Neither (C3) Note: Award (A1) for geometric correct, (A1) for arithmetic correct and (A1) for both “neither”. These may be implied by blanks only if GP and AP correct.

(b) (Series (ii) is a GP with a sum to infinity)

Common ratio 43

(A1)

S∞ = !!!!

"

#

$$$$

%

&

−=

−431

11 ra

(M1)

= 4 (A1) (C3) Note: Do not allow ft from an incorrect series.

[6]

15

36. !!"

#$$%

&!!"

#$$%

&!!"

#$$%

&

710

accept)(2310 37 ax (A1)(A1)(A1)

!!"

#$$%

&

310

= 120 (A1)

120 × 27 a3 = 414 720 (M1) a3 = 27 a = 3 (A1) (C6)

Note: Award (A1)(A1)(A0) for !!"

#$$%

&

310

27 ax3. If this leads to the

answer a = 27, do not award the final (A1). [6]

37. 5 38(2) ( 3 )

3x

! "−$ %

& '

8Accept

5! "! "# $# $

% &% & (M1)(A1)(A1)(A1)

Term is 348384x− (A2) (C6) [6]

38. METHOD 1 2log 2logx x= (A1)

1log log

2y y= (A1)

3log 3logz z= (A1)

12log log 3log

2x y z+ − (A1)(A1)

12 32

a b c+ − (A1) (C6)

16

METHOD 2

2 2 3 3210 , 10 , 10b

a cx y z= = = (A1)(A1)(A1)

2 2 2

10 103 3

10 10log log

10

ba

c

x yz

! "! " ×$ %=$ % $ %$ %

$ %& '& '

(A1)

2 3

210log 10 2 3

2

ba c ba c+ −" # " #= = + −$ % $ %

& '& ' (A2)

[6]

39. (a) (i) $11400, $11800 (A1) 1

(ii) Total salary 10 (2 11000 9 400)2

= × + × (A1)

= $128000 (A1) (N2) 2

(b) (i) $10700, $11449 (A1)(A1)

(ii) 10th year salary 910 000(1.07)= (A1)

= $18384.59 or $18400 or $18385 (A1) (N2) 4

17

(c) EITHER

Scheme A ( )A 2 11000 ( 1)4002nS n= × + − (A1)

Scheme B B10 000(1.07 1)

1.07 1

n

S −=

− (A1)

Solving B AS S> (accept B AS S= , giving 6.33n = ) (may be implied) (M1)

Minimum value of n is 7 years. (A1) (N2)

OR

Using trial and error (M1)

Arturo Bill

6 years $72 000 $71532.91

7 years $85 400 $86 540.21

(A1)(A1) Note: Award (A1) for both values for 6 years, and (A1) for both values for 7 years.

Therefore, minimum number of years is 7. (A1) (N2) 4 [11]

40. (a) Recognizing an AP (M1) u1 =15 d = 2 n = 20 (A1) 4 substituting into u20 = 15 + (20 –1) × 2 M1 = 53 (that is, 53 seats in the 20th row) A1

(b) Substituting into S20= 220 (2(15) + (20–1)2) (or into

220 (15 + 53)) M1

= 680 (that is, 680 seats in total) A1 2 [6]

41. (a) 5000(1.063)n A1 1

(b) Value = $5000(1.063)5 (= $6786.3511...) = $6790 to 3 sf (Accept $6786, or $6786.35) A1 1

18

(c) (i) 5000(1.063)n > 10000 or (1.063)n > 2 A1 1

(ii) Attempting to solve the inequality «log (1.063) > log 2 (M1) n > 11.345... (A1) 12 years A1 3

Note: Candidates are likely to use TABLE or LIST on a GDC to find n. A good way of communicating this is suggested below.

Let y = 1.063x (M1) When x = 11, y = 1.9582, when x = 12, y = 2.0816 (A1) x = 12 ie 12 years A1 3

[6]

42. (a) 1 1 7u S= = (A1) (C1)

(b) 2 2 1 18 7u S u= − = −

11= (A1)

11 7d = − (M1)

4= (A1) (C3)

(c) 4 1 ( 1) 7 3(4)u u n d= + − = + (M1)

4 19u = (A1) (C2) [6]

43. (a) 6 terms (A1) (C1)

(b) ( )23 2510, ( 2) 8,

3x

! "= − = −$ %

& ' (A1)(A1)(A1)

fourth term is 480x− (A1)

for extracting the coefficient 80A = − (A1) (C5) [6]

19

44. METHOD 1

2 39 3 , 27 3= = (A1)(A1)

expressing as a power of 3, 2 2 3 1(3 ) (3 )x x−= (M1)

4 3 33 3x x−= (A1)

4 3 3x x= − (A1)

7 3x =

37

x⇒ = (A1) (C6)

METHOD 2

( )2 log9 1 log27x x= − (M1)(A1)(A1)

2 log27 31 log9 2xx

! "= =# $− & ' (A1)

4 3 3x x= − (A1) 7 3x =

37

x⇒ = (A1) (C6)

Notes: Candidates may use a graphical method. Award (M1)(A1)(A1) for a sketch, (A1) for showing the point of

intersection, (A1) for 0.4285…., and (A1) for 37

.

[6]

45. (a) 3 3 3log log ( 5) log5xx xx

! "− − = $ %−& ' (A1)

5x

Ax

=−

(A1) (C2)

Note: If candidates have an incorrect or no answer to part (a) award (A1)(A0)

if !"

#$%

&− 5

logxx seen in part (b).

20

(b) EITHER

3log 15xx

! " =# $−& '

( )13 35xx

= =−

(M1)(A1)(A1)

3 15x x= −

2 15x− = −

152

x = (A1) (C4)

OR

10

10

log5 1

log 3

xx

! "# $−& ' = (M1)(A1)

10 10log log 35xx

! " =# $−& ' (A1)

7.5x = (A1) (C4) [6]

46. METHOD 1

Using binomial expansion (M1)

( ) ( ) ( ) ( )32233773

23

7313

373 +!!"

#$$%

&+!!

"

#$$%

&+=+ (A1)

= 27 + 27 7 + 63 + 7 7 (A2)

( ) 73490733

+=+ (so p = 90 , q = 34 ) (A1)(A1)(C3)(C3)

METHOD 2

For multiplying (M1)

( ) ( ) ( )( )73776973732

+++=++ (A1)

= 27 + 9 7 + 18 7 + 42 + 21 + 7 7

( = 27 + 27 7 + 63 + 7 7 ) (A2)

( ) 73490733

+=+ (so p = 90 , q = 34 ) (A1)(A1)(C3)(C3) [6]

47. For using u3 = u1r2 = 8 (M1)

8 = 18r2 (A1)

21

r2 = !"

#$%

&=94

188

r = ±32

(A1)(A1)

,11

ruS−

=∞

)8.10(554,54 ==∞S (A1)(A1)(C3)(C3)

[6]

48. (a) (i) logc 15 = logc 3 + logc 5 (A1)

= p + q A1 N2

(ii) logc 25 = 2 logc 5 (A1)

= 2q A1 N2

(b) METHOD 1

621

=d M1

d = 36 A1 N1

METHOD 2

For changing base M1

eg dd 1010

10

10 log6log2,21

log6log

==

d = 36 A1 N1 [6]

49. (a) For taking an appropriate ratio of consecutive terms (M1)

r = 32

A1 N2

(b) For attempting to use the formula for the nth term of a GP (M1)

u15 = 1.39 A1 N2

22

(c) For attempting to use infinite sum formula for a GP (M1)

S = 1215 A1 N2 [6]

50. (a) (i) r = −2 A1 N1

(ii) u15 = −3 (−2)14 (A1)

= −49152 (accept −49200) A1 N2

(b) (i) 2, 6, 18 A1 N1

(ii) r = 3 A1 N1

(c) Setting up equation (or a sketch) M1

182

31

+

+=

−

+

xx

xx

(or correct sketch with relevant information) A1

x2 + 2x + 1 = 2x2 + 2x − 24 (A1)

x2 = 25

x = 5 or x = −5

x = −5 A1 N2 Notes: If “trial and error” is used, work must be documented with several trials shown. Award full marks for a correct answer with this approach. If the work is not documented, award N2 for a correct answer.

(d) (i) r = 21

A1 N1

(ii) For attempting to use infinite sum formula for a GP (M1)

S =

211

8

−

−

S = −16 A1 N2 Note: Award M0A0 if candidates use a value of r where r > 1, or r < −1.

[12]

23

51. (a) (i) S4 = 20 A1 N1

(ii) u1 = 2, d = 2 (A1)

Attempting to use formula for Sn M1

S100 = 10100 A1 N2

(b) (i) M2 = !!"

#$$%

&

1041

A2 N2

(ii) For writing M3 as M2 × M or M × M2 !!"

#$$%

&!!"

#$$%

&!!"

#$$%

&

1041

1021

or M1

M3 = !!"

#$$%

&

++

++

10002401

A2

M3 = !!"

#$$%

&

1061

AG N0

(c) (i) M4 = !!"

#$$%

&

1081

A1 N1

(ii) T4 = !!"

#$$%

&+!!"

#$$%

&+!!"

#$$%

&+!!"

#$$%

&

1081

1061

1041

1021

(M1)

= !!"

#$$%

&

40204

A1A1 N3

(d) T100 = !!"

#$$%

&++!!

"

#$$%

&+!!"

#$$%

&

102001

...1041

1021

(M1)

= !!"

#$$%

&

100010100100

A1A1 N3

[16]

52. (a) ln a3b = 3ln a + ln b (A1)(A1)

ln a3b = 3p + q A1 N3

24

(b) ln 21

=ba ln a − ln b (A1)(A1)

ln 21

=ba p − q A1 N3

[6]

53. Note: Throughout this question, the first and last terms are interchangeable.

(a) For recognizing the arithmetic sequence (M1) u1 = 1, n = 20, u20 = 20 (u1 = 1, n = 20, d = 1) (A1) Evidence of using sum of an AP M1

S20 = ( ) ( ))11912

220

(or220201

×+×=+ S A1

S20 = 210 AG N0

(b) Let there be n cans in bottom row

Evidence of using Sn = 3240 (M1)

eg ( ) ( )( ) ( )( )( ) 3240112

2,324012

2,3240

21

=−−+=−+=+ nnnnnnn

n2 + n − 6480 = 0 A1

n = 80 or n = −81 (A1)

n = 80 A1 N2

25

(c) (i) Evidence of using S = ( )2

1 nn+ (M1)

2S = n2 + n A1

n2 + n − 2S = 0 AG N0

(ii) METHOD 1

Substituting S = 2100

eg n2 + n − 4200 = 0, 2100 = ( )2

1 nn+ A1

EITHER

n = 64.3, n = −65.3 A1

Any valid reason which includes reference to integer being needed, R1

and pointing out that integer not possible here. R1 N1

eg n must be a (positive) integer, this equation does not have integer solutions.

OR

Discriminant = 16 801 A1

Valid reason which includes reference to integer being needed, R1

and pointing out that integer not possible here. R1 N1

eg this discriminant is not a perfect square, therefore no integer solution as needed.

METHOD 2 Trial and error S64 = 2080, S65 = 2145 A1A1

Any valid reason which includes reference to integer being needed, R1 and pointing out that integer not possible here. R1 N1

[14]

54. (a) 51

(0.2) A1 N1

26

(b) (i) 9

10 5125 !"

#$%

&=u (M1)

= 0.0000128 !!

"

#

$$

%

&×!

"

#$%

& −

781251,1028.1,

51 5

7

A1 N2

(ii) 1

5125

−

"#

$%&

'=

n

nu A1 N1

(c) For attempting to use infinite sum formula for a GP !!!!

"

#

$$$$

%

&

!"

#$%

&−511

25 (M1)

S = ( )fs3to3.3125.314125

== A1 N2

[6]

55. (a) loga 10 = loga (5 × 2) (M1)

= loga 5 + loga 2

= p + q A1 N2

(b) loga 8 = loga 23 (M1)

= 3 loga 2

= 3q A1 N2

(c) loga 2.5 = loga 25

(M1)

= loga 5 − loga 2

= p − q A1 N2 [6]

56. (a) 7 terms A1 N1

27

(b) A valid approach (M1)

Correct term chosen ( ) ( )333 336

xx −""#

$%%&

' A1

Calculating ( ) 273,2036 3 −=−=""#

$%%&

' (A1)(A1)

Term is −540x12 A1 N3 [6]

57. Identifying the required term (seen anywhere) M1

eg 22810

×""#

$%%&

'

!!"

#$$%

&

810

= 45 (A1)

4y2, 2 × 2, 4 (A2)

a = 180 A2 N4 [6]

58. (a) For taking three ratios of consecutive terms (M1)

( )3162486

54162

1854

=== A1

hence geometric AG N0

(b) (i) r = 3 (A1)

un = 18 × 3n − 1 A1 N2

(ii) For a valid attempt to solve 18 × 3 n − 1 = 1062882 (M1)

eg trial and error, logs

n = 11 A1 N2 [6]

59. (a) 3, 6, 9 A1 N1

28

(b) (i) Evidence of using the sum of an AP M1

eg ( ) 312032220

×−+×

∑=

=20

1

6303n

n A1 N1

(ii) METHOD 1

Correct calculation for ∑=

100

1

3n

n (A1)

eg ( ) 15150,399322100

×+×

Evidence of subtraction (M1)

eg 15150 − 630

145203100

21

=∑=n

n A1 N2

METHOD 2

Recognising that first term is 63, the number of terms is 80 (A1)(A1)

eg ( ) ( )379126280,30063

280

×++

145203100

21

=∑=n

n A1 N2

[6]

29

60. (a) For finding second, third and fourth terms correctly (A1)(A1)(A1)

Second term ,e1e

14 3 !

"

#$%

&!!"

#$$%

& third term ,

e1e

14 2

2 !"

#$%

&!!"

#$$%

&

fourth term 3

e1e

14

!"

#$%

&!!"

#$$%

&

For finding first and last terms, and adding them to their three terms (A1)

+!"

#$%

&!!"

#$$%

&+!

"

#$%

&!!"

#$$%

&+!"

#$%

&!!"

#$$%

&+!!

"

#$$%

&=!

"

#$%

&+

32234

4

e1e

34

e1e

24

e1e

14

e04

e1e

4

e1

44

!"

#$%

&!!"

#$$%

&

432

2344

e1

e1e4

e1e6

e1e4e

e1e !

"

#$%

&+!

"

#$%

&+!

"

#$%

&+!"

#$%

&+=!

"

#$%

&+

!"

#$%

&++++= 42

24

e1

e46e4e N4

(b) 432

2344

e1

e1e4

e1e6

e1e4e

e1e !

"

#$%

&+!

"

#$%

&−!

"

#$%

&+!"

#$%

&−=!

"

#$%

&−

!"

#$%

&+−+−= 42

24

e1

e46e4e (A1)

Adding gives 2e4 + 12 + 4e2

!!

"

#

$$

%

&!"

#$%

&!!"

#$$%

&+!

"

#$%

&!!"

#$$%

&+!!

"

#$$%

& 4224

e1

44

2e1e

24

2e04

2accept A1 N2

[6]

61. (a) d = 3 (A1) evidence of substitution into un = a + (n − 1) d (M1) eg u101 = 2 + 100 × 3 u101 = 302 A1 N3

30

(b) correct approach (M1) eg 152 = 2 + (n − 1) × 3 correct simplification (A1) eg 150 = (n − 1) × 3, 50 = n − 1, 152 = −1 + 3n n = 51 A1 N2

[6]

62. (a) evidence of dividing two terms (M1)

eg 10801800,

30001800

−−

r = − 0.6 A1 N2

(b) evidence of substituting into the formula for the 10th term (M1)

eg u10 = 3000(− 0.6)9

u10 = −30.2 (accept the exact value −30.233088) A1 N2

(c) evidence of substituting into the formula for the infinite sum (M1)

6.13000.. =Sge

S = 1875 A1 N2 [6]

63. evidence of using binomial expansion (M1)

eg selecting correct term, ...28

18 26708 +!!

"

#$$%

&+!!

"

#$$%

&+ bababa

evidence of calculating the factors, in any order A1A1A1

eg 56, ( )53

53

3

332

58

,3,32

−"#

$%&

'""#

$%%&

'− x

−4032x3 (accept = −4030x 3 to 3 sf) A1 N2 [5]