2012 Y5 Assig 4 AP GP (Solution)

Raffles Institution H2 Mathematics 9740 Year 5 2012

_________________________ Assignment 4 : Arithmetic and Geometric Progressions 1

N91/1/12a a = 1 and d = 3 : un = 1000 = 1 + 3(n − 1) ⇒ n = 334 Hence S334 =

334 (1 + 1000) = 167 167 . 2

Consider the AP formed by every third term of the progression: 7, 16, 25, …, This new AP has a = 7 and d = 9 .

1 3

To find the last term of this new AP : 7 + 9( n − 1) < 1000 ⇒ n < 111 ⇒ n = 111 Hence, sum of the remaining terms = 167 167 − 2

111 [2(7) + 110(9)] = 167167 − 55722 = 111 445 2

06/SAJC/H2/7a

un = S n − S n−1 = [5n + n 2 ] − [5(n − 1) + (n − 1)2 ] = 5n + n 2 − 5n + 5 − n 2 + 2n − 1 = 2n + 4 Since un − un−1 = 2n + 4 − 2(n − 1) − 4 = 2 is a constant, the series form an arithmetic progression.

u1 = 6 , d = 4 and n = 100 : S100 = 50[2(6) + 99(4)] = 20 400 . 3

N2001/1/14

a a = 16 and S B = = 64 . Since rB = (rA ) 2 , we have 1 − rA 1 − rB a a 3 = 64 ⇒ = 64 ⇒ 16 = 64(1 + rA ) ⇒ rA = − . 2 (1 − rA )(1 + rA ) 4 1 − (rA )

(a) Consider S A =



 3   4 

Hence a = 16  1 −  −   = 28 .



(b) u9 = 43 ⇒ a + 8d = 43 ⇒ a = 43 − 8d Hence

S15 = 570 ⇒

15 (2a + 14d ) = 570 ⇒ 2(43 − 8d ) + 14d = 76 ⇒ 10 = 2d ⇒ d = 5 ⇒ a = 3 2

So

n Sn > 2265 ⇒ [2(3) + 5(n − 1)] > 2265 ⇒ 5n2 + n − 4530 > 0 ⇒ (5n + 151)(n − 30) > 0 2 151 and we have n > 30 or n < − . Thus the least value of n = 31 . 5 ___________________________________________ Assignment 4 : Arithmetic and Geometric Progressions Page 1 of 6

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4

06/TJC/Promos/6 Given that 2T5 = S2, we have i.e.,

2ar4 = a + ar, 2r4 – r – 1 = 0.

From G.C., r = −0.647799 = - 0 .648 (3 s.f) or r = 1. (rejected as |r| < 1 for S to exist). For Sn is within 5% of S, i.e., |Sn – S| < 0.05S , we have

Since

a (1 − r n ) a 0.05a − < 1− r 1− r 1− r

--------------(1)

a > 0 (as a > 0 and r = −0.647799 ), (1) simplifies to 1− r (1 − (−0.647799) n ) − 1 < 0.05

(0.647799)n < 0.05 n lg(0.647799) < lg ( 0.05) n>

lg(0.05) = 6.90 (correct to 3 significant figures) lg(0.647799)

Hence, minimum value of n is 7.

5

N93/1/15(a)

a S ar ar Sr 1 , we have − = = = ⇒ 1 + r = −2r ⇒ r = − . 2 1− r 2 1− r (1 − r )(1 + r ) 1 + r 3 a 18 81 (i) u3 = ar 2 = 2 ⇒ a = 18 , and so required sum to infinity = = = . 1 − r 2 1 − 19 4 1 18 18 (ii) rH = ⇒ S H = = 27 . Since S = = 13.5 , S H = 2S . 3 1 − 1/ 3 1 + 1/ 3 Since S =

6

06/MI/Prelims/P2/1

a + 4d a + 10d a = ⇒ (a + 4d ) 2 = a (a + 10d ) ⇒ d (8d − a ) = 0 ⇒ d = 0 or d = a a + 4d 8 a + 4d Now, d = 0 ⇒ r = = 1. a a 20 + 10 Also, from S20 = 10(2a + 19d ) = 875 , d = ⇒ a = 20 ⇒ r = >1 8 20 Since | r | is not less than 1 for both cases, the sum to infinity does not exist. 7

N2005/2/4

b c a+b = ⇒ b 2 = ac and c − a = b − c ⇒ 2c = a + b ⇒ c = a b 2 2 b   b2  b b a+b 2 2 (ii) b 2 = ac ⇒ b 2 = a   ⇒ 2b = a + ab ⇒ 2  2  = 1 + ⇒ 2  2  − − 1 = 0 a  2  a  a  a

(i)

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 b2 2 a

 b   b   b  b b 1  − − 1 = 0 ⇒  2   + 1  − 1 = 0 ⇒ = 1 or = − a a 2   a   a   a b 1 a 2a Since the sum to infinity exists, = − , and so S = = . b a 2 3 1− a

(iii) 2 

8

(i) u25 = (16000)(1.04) 24 = 41012.87 . He will earn $41 012.87 in his 25th year.

16000(1.0425 − 1) (ii) S 25 = = 666334.53 1.04 − 1 He earns $666 334.53 over the 25-year period.

16000(1.04n − 1) > 1000000 ⇒ 1.04n − 1 > 2.5 1.04 − 1 ⇒ 1.04n > 3.5 (iii) ⇒ n

> 31.9

Minimum number of years he has to work = 32 9

06/AJC/H2/5 (a)

Number of weeks needed to sell off all the chickens =

1000 k

Total Cost = ( 0.5 ) (1000 ) + (1000 − k ) + (1000 − 2k ) +
+ ( k ) 

 1  1000    [1000 + k ] 2  k  

= ( 0.5 )   =$ (b)

250 (1000 + k ) k

(shown)

When k = 40, number of weeks =

1000 = 25 40

2 24 Total Earnings = 12 ( 40 ) 1 + ( 0.95 ) + ( 0.95 ) +
+ ( 0.95 ) 





1 − ( 0.95 )    1 − ( 0.95 )  25

= 480 

= $6937.06 ∴ Total Profits = Total Earnings – Total Cost = 6937.06 –

250 (1000 + 40 ) 40

= 6937.06 – 6500 = $437

(to 3 s.f.)

∴ The farmer made a profit.

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10

06/NJC/JC1 Common Test/11 (i) Time taken on nth day = 45(0.985) n −1.

45(0.985) n −1 < 30 ⇒ ( n − 1) ln(0.985) < ln

⇒ n − 1 > ln

30 45

30 ÷ ln(0.985) 45

⇒ n > 27.8 ∴ least number of days =28. (ii) From (i), Tony has to train daily for 28 days to obtain the ‘Running’ award.

Total time spent running over 28 days 28

= ∑ 45(0.985)n −1 n =1

45(1 − 0.98528 ) 1 − 0.985 = 1035.12 mins (>15 hrs = 900 mins) =

Thus, Tony is putting himself at risk of having asthma. 11

N07/P1/10 (i)

Given: first term of the geometric series = a, ar = a + 3d and ar 2 = a + 5d .

∴ a (r −1) = 3d

(1) a ( r 2 −1) = 5d

(2) (2) r 2 −1 5 : = , r ≠ 1 since d ≠ 0 from (1) (1) r −1 3 ⇒ 5r − 5 = 3r 2 − 3 ⇒ 3r 2 − 5r + 2 = 0 (Shown) (ii)

3r 2 − 5r + 2 = 0 (3r − 2)( r −1) = 0 2 r = 1 or r = 3 2 Since r ≠ 1 , r = . 3 Since | r | < 1 , the geometric series is convergent. (deduced) a a ∴ S∞ = = = 3a 1 − r 1 − 23

________________________________________ Assignment 4 : Arithmetic and Geometric Progressions Page 4 of 6

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(iii)

n S = [ 2a + (n −1)d ] > 4a 2 2 1 1 Substitute r = into (1), − a = 3d ⇒ d = − a 3 3 9  n 1 So  2a − a ( n −1) > 4a 2  9  1 2an − an(n −1) > 8a 9 1 Since a > 0 , 2n − n( n −1) > 8 9 18n − n 2 + n > 72 n 2 −19n + 72 < 0 From GC, or 5.2280 < n < 13.772 Solution set is { n ∈ + : 6 ≤ n ≤ 13 } .

12

06/ACJC/H2 J1 Promos/7 Let first term and common difference of series A be a and d, and first term and common ratio of series G be b and r.

a+b = −1

(1)

a + d + br = 0

(2)

2

Sub r = 2a, d = 4b into (2),

a + 4b + 2ab = 0 (3) From (1), b = − 1 − a 2

a + 4(− 1 − a) + 2a(− 1 − a) = 0 2

2

2

2a + 4a + 2 = 0 2(a + 1)2 = 0 ∴ a = −1

13

N11/P1/9 (i) Let an be the depth drilled using machine A on day n, and let N be the last day of drilling. Then a1 , a2 , a3 , … , an , … , aN forms an AP with first term a1 = 256 and common difference d = −7. Depth drilled on 10th day is a10 = a1 + (10 − 1)d = 256 + 9(−7) = 193. Day N is the first time depth drilled is below 10 metres, so N is the smallest integer such that aN = 256 + ( N − 1)(−7) < 10. Therefore N >

246 + 1 = 36.143, 7

giving N = 37. Total depth drilled is S N =

37 [2(256) + (37 − 1)(−7)] = 4810 metres. 2 ________________________________________ Assignment 4 : Arithmetic and Geometric Progressions Page 5 of 6

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(ii)

Let bn be the depth drilled using machine B on day n. Then b1 , b2 , b3 , … , bn , … forms a GP with first term b1 = 256 and common ration r = 89 . We want the least 99 S∞ , i.e. n so that Sn ≥ 100 n

99 99 8 S n = (1 − r ) S ∞ ≥ S∞ ⇒ 1 −   ≥ 100  9  100 ln(1 / 100) ⇒ n≥ = 39.1. ln(8 / 9) Therefore the least n is 40, and it takes 40 days for the depth drilled to exceed n

99% of the theoretical maximum depth.

END OF ASSIGNMENT

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