JC2 Revision Package 3 H2 Mathematics (9740) AP & GP Section A

H2 Revision Package 3: AP & GP

2012 Meridian Junior College

JC2 Revision Package 3 H2 Mathematics (9740) AP & GP Section A

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Qn 2 2011 DHS Prelim/P1/1 n

∑ (2r + 1)(2r + 3) r =1 n

= ∑ (4r 2 + 8r + 3) r =1 n

n

r =1

r =1

= 4∑ r 2 + ∑ (8r + 3) n  n = 4  (n + 1)(2n + 1)  + (11 + 8n + 3) 6  2 n 2n = (n + 1)(2n + 1) + ( 8n + 14 ) 3 2 n 3  =  2(2n 2 + 3n + 1) + (8n + 14)  3 2  n = ( 4n 2 + 18n + 23) (shown) 3 75

∑ (2r + 1)(2r + 3) r = 25 75

24

r =1

r =1

= ∑ (2r + 1)(2r + 3) − ∑ (2r + 1)(2r + 3) 75 ( 4(75)2 + 18(75) + 23) + 243 ( 4(24)2 + 18(24) + 23) 3 = 596825 − 22072 = 574753 =

Qn 3 2011 Prelim HCI/P1/5 (i) yn +1 xn +1 + 2

yn

=

xn + 2 1 xn − 1 + 2 1 x +2 1 2 = =  n  = = constant xn + 2 2  xn + 2  2

y1 , y2 , y3 , … is a geometric sequence with common ratio

1 . 2

y1 = x1 + 2 = 3 1 2

n −1

Therefore, yn = 3  

1 xn = yn − 2 = 3   2

.

n −1

−2

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(ii)


n −1

1 3  − 2 xn 1 x  1 2 = n  n  = n   n −1 n 2 yn 2  yn  2 1 3  2 1  2 n −1  1 1 = n 1 − ( 2 )  = n − 2  3  2 3 1 When n → ∞ , n decreases to 0, 2 xn 1 ⇒ n decreases to − 2 yn 3

Qn 4 2011 MI Prelim/P1/13

(a)

Tn = Sn − Sn −1

(

2

= ( 2n 2 + kn − 3) − 2 ( n − 1) + k ( n − 1) − 3

)

= 2n 2 + kn − 3 − ( 2n 2 − 4n + 2 + kn − k − 3) = 4n + k − 2 Tn − Tn −1 = ( 4n + k − 2 ) − ( 4 ( n − 1) + k − 2 ) =4 = constant Hence, sequence is an AP.

(b) (i) A : 5000, 5000 − 50 (1) ,5000 − 50 (1 + 2 ) ,...... Tn = 5000 − 50 (1 + 2 + 3 + ...... + ( n − 1) ) n −1 (n) 2 = −25n 2 + 25n + 5000 = 5000 − 50i

(ii) B : a = 50, r = 1.25

∴Tn = 50 (1.25 ) 50 (1.25 )

n −1

n −1

> −25n 2 + 25n + 5000

Using GC, n > 13.4 ∴ n ≥ 14 ∴ First month of B outperforming A is Feb 2011.

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Qn 5 2011 MJC Prelim/P1/4

(i) T3, T7, T15 are the first three terms of the GP : T T ⇒ r = 15 = 7 T7 T3 ⇒r=

a + 14d a + 6d = a + 6d a + 2d

( a + 14d )( a + 2d ) = ( a + 6d )

2

a 2 + 16ad + 28d 2 = a 2 + 12ad + 36d 2 2d 2 − ad = 0 d ( 2d − a ) = 0 d=

a or 2

d = 0 (rejected ∵ d ≠ 0)

a a + 6  2 =2 ∴r = a a + 2  2 ∵ r = 2 < 1 , the geometric series is NOT convergent / divergent.

(ii) New Sequence:

T2 , T4 , T6 ,... that is, a + d , a + 3d , a + 5d ,...

n  2(a + d ) + (n − 1) ( 2d )  ≤ 24a 2  n  a 2  a +  + ( n − 1)( a )  ≤ 24a  2  2 

S=

 n  3 2   + ( n − 1)(1)  ≤ 24  2 2 

(∵ a > 0 )

n [ n + 2] ≤ 24 2 n 2 + 2n − 48 ≤ 0 −8 ≤ n ≤ 6 ∴1 ≤ n ≤ 6

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Qn 6 2011 NJC Prelim/P1/6

(i) Under Plan A, Year

Start ($)

End ($)

1

2000

2000 + 100

2

2000 + 100

2000 + 100

+2400

+2400 + 100

3

2000 + 100 +2400 + 100 +2800

2000 + 100 +2400 + 100 +2800 + 100

…

…

…

n

2000 + 100 +2400 + 100 +2800 + ...

2000 + 100 +2400 + 100 +2800 + ...

+2000 + ( n − 1) 400

+2000 + ( n − 1) 400 + 100

At the end of n years, Mr. Wei would have accumulated

2000 + 100 + 2400 + 100 + 2800 + ... + 2000 + ( n − 1) 400 + 100 = 2000 + 2400 + 2800 + ... + 2000 + ( n − 1) 400 + 100n n  2 × 2000 + 400 ( n − 1)  + 100n 2 = 2000n + 200n ( n − 1) + 100n =

= 2100n + 200n ( n − 1) (shown) (ii)

2100n + 200n ( n − 1) ≥ 100000 200n 2 + 2100n − 200n − 100000 ≥ 0 n 2 + 9.5n − 500 ≥ 0 Using G.C., n ≤ −27.6 (3 s.f.) or n ≥ 18.1 (3 s.f.) Since n > 0, the least number of years is 19.

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(iii) Let $ x be the amount Ms. Goon invests into her account each year. Under Plan B, Year

Start ($)

End ($)

1

x

1.03x

2

1.03x + x

1.032 x + 1.03 x

3

1.032 x + 1.03 x + x

1.033 x + 1.032 x + 1.03 x

…

…

…

11

1.0310 x + 1.039 x +...

1.0311 x + 1.0310 x

+1.03x + x

+1.032 x + 1.03x

+...

To save at least $60000, we need: x (1.0311 + 1.0310 + ... + 1.032 + 1.03) ≥ 60000 1.03 (1.0311 − 1)   ≥ 60000 x 1.03 − 1   x ≥ 4548.20 Thus, Ms. Goon needs to add at least $4549 (nearest dollar) to her account each year. Qn 7 2011 PJC Prelim/P1/3

(i)

Sn 2

3

n

 3 3 3 3 = 4 + 2   4 + 2   4 + 2   4 + ...... + 2   4  4 4 4 4 2 3 n  3   3   3  3  = 4 + 2 ( 4 )   +   +   + ...... +     4    4   4   4    3 n  1 −    H  3    4   = 4 + 8   4  1−  3    4   3 n  = 4 + 24 1 −      4   3 = 28 − 24   4

n

1st

2nd

3rd

(n+1)

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(ii)


n

3 S n = 28 − 24   > 24 4 n

1 3 >  6 4 1 ln   6 n>   3 ln   4 n > 6.23 ⇒ n = 7 The ball must bounce at least 7 times for it to travel more than 24 m.

(iii)

n

3 n → ∞,   → 0, S∞ = 28 4 Since sum to infinity is 28, the ball will not travel more than 28m.

Qn 8 2011 SAJC Prelim/P2/4

(a)

x2 − 3 x2 − 2x + 1 = x2 + 2 x −1 x2 − 3 2 2 ⇒ ( x − 3) = ( x 2 + (2 x − 1))( x 2 − (2 x − 1)) ⇒ x 4 − 6 x 2 + 9 = ( x 4 − (2 x − 1) 2 ) ⇒ −2 x 2 − 4 x + 10 = 0 ⇒ x2 + 2 x − 5 = 0 −2 ± 4 − 4(1)(−5) 2 ∴ x = −1 − 6 or − 1 + 6 ⇒x=

If x = −1 + 6, r = 2.23 (rejected as the sequence converges) If x = −1 − 6, r = −0.225 The common ratio, r = - 0.225 (3 sig figs). (b)

Sapling No 1 2 3 4 … 200 201

Distance walked by farmer 0 2(1) 2(2) 2(3) … 2(199) 200

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Distance covered followed an A.P with common difference 2. Therefore, total distance = (2+4+…+398)+200 = 2(1+2+…+199)+200 199 = 2( (1 + 199) ) +200 2 = 40000 m. (shown) Sapling No 1,2 3,4 5,6 … 199,200 201

Distance walked by farmer 2(1) 2(3) 2(5) … 2(199) 200

Distance covered followed an A.P with common difference 4. Therefore, total distance = (2+6+10+…+199) + 200 = 2(1+3+5+…+199)+ 200 100 = 2( (1 + 199) ) + 200 2 = 20200 m. Hence distance saved by the famer by using the 2nd method is (40000 – 20200)m = 19800 m. Qn 9 2011 SRJC Prelim/P1/12

(a)

4 3 U n = S n − S n −1 4 4   = 5 + n +1 −  5 + n  3 3   4 (1 − 3) −8 = = n +1 3n +1 3 n Un −8 3 = n +1 × U n −1 3 −8 1 = ( constant ) 3 Sn = 5 +

n +1

Hence, the series is a geometric series with common ratio

1 . 3

(b) (i) Let N be the total number of terms in the first n sets. N = 1 + 2 + 3 + ... + n n ( n + 1) = 2 Sum of all numbers in the first n sets: 1 n ( n + 1)   n ( n + 1)   SN = × 2(3) +  −1 × 4   2 2 2    

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=


1 n ( n + 1) ( n 2 + n + 1) 2

(ii) The last term in the nth set  n ( n + 1)  −1 × 4 UN = 3+  2   2 = 2n + 2 n − 1 Hence the first term in the nth set = the last term in the (n-1)th set + 2 =  2( n − 1) 2 + 2( n − 1) − 1 + 4 = 2n 2 − 2n + 3 The sum of the numbers in the nth set n = ×  ( 2n 2 + 2n −1) + ( 2n 2 − 2n + 3)  2 = n ( 2n 2 + 1)

(iii) For n ( 2n 2 + 1) > 4500 From G.C, n > 13.09. Thus least n is 14.

Qn 10 2011 TPJC Prelim/P1/3

Tn = log 2xn-1 = log 2 + (n-1) log x Tn-1 = log 2 + (n − 2) log x Tn - Tn-1 = log x which is independent of n Thus, Tn forms an arithmetic progression with a = log 2 and d = log x First term of GP = log 2 + 9 log x – log 2 = 9 log x Second term of GP = log 2 + 3 log x – log 2 = 3 log x Third term of GP = log 2 + log x – log 2 = log x Hence, r =

1 3

9 log x 1 1− 3 27 20log 2 + 190 log x > log x 2 log x > − 0.034111 x > 0.924

10[2 log 2 + 19log x] >

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Qn 11 2011 TJC Prelim/P1/9

(a) Sum of all integers between 1102 and 2011 inclusive 2011 − 1102 + 1 = (1102 + 2011) = 1416415 2 1104, 1107, …, 2010 is an AP with first term 1104 and common difference 3 2010 = 1104 + ( n –1) 3 ⇒ n = 303 Sum of integers between 1102 and 2011 that are divisible by 3 303 = (1104 + 2010 ) 2 = 471771 ∴ required sum = 1416415 – 471771 = 944644

(b) Let Gn and Hn be the nth term of G and H respectively. 2

2 ( Gn ) = ( ar n −1 ) = r 2 , a constant which is independent of n Hn = H n −1 ( Gn −1 )2 ( ar n − 2 )2

Therefore, H is a geometric progression. a2  a  = 2  2 1− r  1− r 

(1 − r )

2

2

= 2 (1 − r 2 )

3r 2 − 2r − 1 = 0 1 r = − or r = 1 (rejected since r < 1 ) 3 Qn 12 2011 YJC Prelim/P1/7

(i) Let the length of a side of S 2 be a2 . 2

2

 a1   a1  2   +   = a2 2 2 a a2 = 1 2

Geometric sequence with first term = a 1 and common ratio =

 1  Length of a side of S3 = a1    2 (ii)

 1  Length of a side of Sn = a1    2

3−1

=

1 . 2

a1 2

n −1

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2

  1 n−1  Area of S n , Tn =  a1    =   2   

( 2)

2n−2

2

a = n1−1 2

a12 a12 = . 2n +1−1 2n

Area of S n +1 , Tn +1 =

Tn+1 Tn

2

a1

a12 n 1 = 2 2 = = a constant a1 2 n −1 2

Therefore, the areas of consecutive squares form a GP with common ratio Geometric progression with first term = a12 and common ratio =

1 . 2

1 . 2

2

Sum to infinity =

a1

1 1− 2

2

= 2a1

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