H2 Revision Package 3: AP & GP. 2012 Meridian Junior College. Page 1 of 11.
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
Page 1 of 11
H2 Revision Package 3: AP & GP
2012 Meridian Junior College
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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H2 Revision Package 3: AP & GP
(ii)
2012 Meridian Junior College
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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H2 Revision Package 3: AP & GP
2012 Meridian Junior College
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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H2 Revision Package 3: AP & GP
2012 Meridian Junior College
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.
Page 5 of 11
H2 Revision Package 3: AP & GP
2012 Meridian Junior College
(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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H2 Revision Package 3: AP & GP
(ii)
2012 Meridian Junior College
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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H2 Revision Package 3: AP & GP
2012 Meridian Junior College
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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H2 Revision Package 3: AP & GP
=
2012 Meridian Junior College
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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H2 Revision Package 3: AP & GP
2012 Meridian Junior College
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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H2 Revision Package 3: AP & GP
2012 Meridian Junior College
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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