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1.1
Warm-up Practice:
A.
1. 1, 3, 5, 7; 19; 29 2. 5, 8, 11, 14; 32; 47
3. 3, 6, 11, 18; 102; 227 4. -2, 1, 6, 13; 97; 222
5. , 2/3, , 4/5; 10/11; 15/16 6. 1/5, 1/3, 3/7, 1/2; 5/7; 15/19
7. 2, 3/2, 4/3, 5/4; 11/10; 16/15 8. 0, , 2/3, ; 9/10; 14/15
9. 0, , 2/5, ; ; 14/17 10. , 4/5, 5/6, 6/7; 12/13; 17/18
11. -1, 2, -3, 4; 10; -15 12. -1, 2, -3, 4; 10; -15
B.
13. 32 14. 33
15. 208 16. 9/10
C.
17. An = 2n 1 18. An = 2n
19. An =
; An = 1 +
20. An =
An =
21. An = 3n 2 22. An =
; An = 1
23. An = 2n x (-1)n - 1
;An = 2n x (-1)n + 1
24. An = 3n x (-1)^(n 1);An = 3n x (-1)n + 1
25. An = -3n + 2 26. An = -4n + 3
27. An = (5)n; 5
n/2 28. An = n
2 + n
D.
29. 63 30. 126
31. 43/10 or 4 3/10 32. 17/4, 4 1/4
33. -13/15 34. 5/12
E.
35. 2 + 4 + 6 + 8 + 10 + 12 = 42 36. 1 + 4 + 7 + 10 + 13 + 16 + 19 + 22 = 92
37. 0 + + 2/3 + + 4/5 = 163/60; 2 43/60 38. 1 + 1 + 1 + 1 +1 = 5
39. -1 + 1 1 + 1 1 + 1 = 0 40. + 1/3 + + 1/5 = 77/60;1 17/60
F.
-
41. 2x, 4x, 6x, 8x, 10x 42. x, x2, x
3, x
4, x
5, x
6
43. 1/x3, 1/x
4, 1/x
5, 1/x
6, 1/x
7, 1/x
8 44. x, 2x
2, 3x
3, 4x
4, 5x
5
45. 1/x, 4/x2, 9/x
3, 16/x
4, 25/x
5 46.
,
,
,
,
Powerplus
A.
1. (x + 1) + (x + 2) + (x + 3) + (x + 4) 2. (x 3) + (x 4) + (x 5) + (x 6)
3. (x + 1) + (x + 1) + (x + 1)4 + (x + 1)
5 + (x + 1)
6 + (x + 1)
7 4. (x + 3) + (x + 3)
2 + (x + 3)
3 + (x + 3)
4
5.
+
+
+
+
6.
+
+
+
+
+
B.
1. 2. 3.
4.
5.
6.
7.
8.
9.
10.
11. 12. 13.
14.
C.
2 is True. (walang page 14, na ulit yung 13)
D.
a.
n F1 + F2 + . Fn F12 + F2
2 + . Fn
2 Fn + 2 1 FnFn + 1
1 1 1 1 1
2 2 2 2 2
3 4 6 4 6
4 7 15 7 15
5 12 40 12 40
Notice that:
1. F1 + F2 + . Fn = Fn + 2 1
2. F12 + F2
2 + . Fn
2 = FnFn + 1
-
b.
n F2
n + 1
FnFn + 2 + (-1)n
1 1 1
2 4 4
3 9 9
4 25 25
5 64 64
Yes.
Proof by Induction:
First, for n = 1, . F2
2 = F1F3 + (-1)n
=> 12 = 1(2) 1. This is true.
For n = 2, 22
= 1(3) + 1. Also True.
Next, assume that 1 F2
n + 1 = FnFn + 2 1 is true for some n that is a positive odd integer.
We know that Fn = Fn + 2 Fn + 1 . Subsitute, we get
F2
n + 1 = ( Fn + 2 Fn + 1) Fn + 2 1
F2
n + 1 = F2
n + 2 - Fn + 1Fn + 2 1
F2
n + 2 = F2
n + 1 + Fn + 1Fn + 2 + 1
F2
n + 2 = Fn + 1(Fn + 1 + Fn + 2) + 1
But, Fn + 1 + Fn + 2 = Fn + 3 , so,
2F2
n + 2 = Fn + 1Fn + 3 + 1. So, 1 implies 2. Let n + 1 = a, where a is even, since n is odd. Similarly,
F2
a + 2 = ( Fa + 3 Fa + 2) Fa + 3 + 1
F2
a + 2 = F2
a + 3 - Fa + 2Fa + 3 + 1
F2
a + 2 = F2
a + 3 + Fa + 2Fa + 3 + 1
F2
a + 3 = Fa + 2(Fa + 2 + Fa + 3) 1, but Fa + 2 + Fa + 3 = Fn + 4 .
3. F2
a + 2 = FaFa + 3 2 1. => F2
n + 3 = Fn + 2Fn + 4 1. We know that n + 3 is even. Therefore, the equation will have + 1 when n is
even and 1 when n is odd. We can denote that as:
F2
n + 1 = FnFn + 2 + ( 1)n , where n is a positive integer.
Take the Challenge:
1. 5. (37 + 38, 24 + 25 + 26, 13 + 14 + 15 + 16 + 17, 10 + 11 + 12 + 13 + 14 + 15,3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12)
2. 7747
3. 1
4. 5
5. 10
-
1.2
A.
1. Arithmetic; d = 1 2. Arithmetic; d = 3
3. Not Arithmetic 4. Arithmetic; d = 4
5. Arithmetic; d = 5 6. Not Arithmetic
7. Not Arithmetic 8. Arithmetic; d =
9. Arithmetic; d = 2/3 10. Not Arithmetic
B.
11. an = 5n 3 12. an = 3n + 8
13. an =
14. an = 10n 6
15. an = 3n 6 16. an = 4n 9.4721
C.
17. 76 18. 9/2
19. 16.207 20. 43/5
21. 1 22. 37
23. 5 24. 8.4
D.
25. 16 26. 15
27. 46 28. 13
29. 21 30. 14
E.
31. 10 32. 21
33. 96 34. 44
35. 1220 36. 210
37. 3775 38. 765
F.
42, 44, 46 are true.
-
Powerplus
A.
1. 8, 13, 18; 88; Sum = 816 2. 2, 8, 14; 116; 1180
3. 2, -1, -4; -85; Sum = -1245 4. 4, 2, 0; -74; -1400
5. 4, 8, 12; 400; Sum = 20,200 6. -2, -4, -6; -200; Sum = -10100
B.
1. 396 2. 504
3. 1275 4. 3488
5. -27 6. -65
7. 3888 8. 320
9. 7 10. 630 cm
11. No. The last row will contain 3 blocks, and there will be a total of 9 rows.
12. 6 13. P7100 14. P465
15.
a.
b.
y(x + z) = 2xz y(x + z) = 2xz
xy + yz = 2xz
xy xz = xz yz
x(y z) = z(x y)
16. Kulang sa info.
17. 579 18. 99
19.
20. Sn =
=
= 3n
2 n
Cancel n on both sides, we get
= 3n 1
2 + an = 6n 2
an = 6n 4
-
Take the Challenge
1. 40
2. 78
3. 92
4. 166 ( not sure )
5. 45
SECTION 1.3
Warmup Practice
A.
1. Geometric; ratio = 5 2. Geometric; ratio = 2
3. Geometric; ratio = 1/3 4. Not geometric
5. Not geometric 6. Geometric; ratio = -1/3
7. Geometric; ratio = -2 8. Not geometric
9. Geometric; ratio = 1/b 10. Geometric; ratio = x/2
B.
11. 10, 5, 5/2, 5/4, 5/8 12. 12, 4, 4/3, 4/9, 4/27
13. -1/4, , -1, 2, -4 14. -1/16, , -1, 4, -16
15. 3, -9, 27, -81, 243 16. 5, -25, 125, -625, 3125
17.
, 2a, 4b,
18. c, b,
,
,
C.
19. 3, 6, 12, 24, 48 20. -2, 4, -8, 16, -32
21. -5, -1, -1/25, -1/125,-1/625 22. 8, 4, 2, 1,
23. 2, 8, 32, 128, 512 24. 3, 6, 12, 24, 48
25. -3, -12, -36, -192, -768 26. 2, -10, 50, -250, 1250
27. -64, 32, -16, 8, -4 28. -64, -32, -16, -8, -4
29. -64, -32, -16, -8, -4 30. -81, 27, -9, 3, -1
31. 2, 10, 50, 250, 1250 32. -3, 9, -27, 81, -243
D.
33. 728 34. 381
35. 33 36. -728
-
37. -64.5 38. -85/24 or -3 13/24
E.
39. 1093 40. 55987
41. 547 42. 171
43. 31 44.33/32 or 1 31/32
45. 26/81 46. 2343/1024 or 2 295/1024
Powerplus
A.
1. 1/16 2. -1/125
3. -1 4. -1
5. 32; 62 + 6. 81; 121 +
7. 2 8.
9. 3584 10. 1/32
11. 1/27 12.
13. -3 14.
15. 6, 18, 54; -6, 18, -54 16. 6, 12, 24, 48
17. -16 18. 27
19. Impossible 20. 5
B.
1.
sq. units 2. 2047
3. 163.604 or 40951/250 4. Approximately 1286.831681 m
Take the Challenge
1. 28.
2. First, we know that p =
=
=
. Also, the sum of the integers from 1 to p is equal to
=
=
. By the difference of squares property, we have:
=
. Next, we
have 1 + 9 + 92 + . + 9
n =
=
=
. But, 9
n + 1 = 3
2n + 2 .So, we have:
=
3. -516096
4. a. First, we know that r% = 0.01r. Also,let T0 be the starting value of the quantity. So, r% of T0 = 0.01r x T0. So, T1 = T0 +
0.01r x T0 = T0(1 + 0.01r). Similarly, we have T2 = T1(1 + 0.01r) = T0(1 + 0.01r)2. So, this means that T1 , T2 , T3 , . Tn
form a geometric sequence with common ratio (1 + 0.01r) and Tn = T0(1 + 0.01r)n .
b. 8,144,473.134.
-
5. 1139
6. No integral value of x.
Section 1.4
Warmup Practice
A.
1. 24 2. 4
3. No sum 4. 3
5. 1/4 6. No sum
7. 8/49 8. 8/7
9. 2 10. -1/7
B.
11. 1/3 12.2/3
13. 31/99 14. 5/11
15. 2 410/999 16.3 1/37
17. 29/225 18. 5/6
C.
19. 20 cm 20.120 cm
21. 30 cm 22. 24 cm
Powerplus
A.
1. -1/2 2.
3. a1 = 1; r = 1/3 4. 21/8
B.
1. d3 = 7.2; d5 = 6.48; d2n 1 = 8 x 0.9(n 1)
2. d4 = 8.1; d2n = 10 x 0.9(n 1)
3. a. 80(1 - 0.9n) b. 100(1 0.9
n)
4. 190
C.
1. a. 1/2n
b. 10 cuts
-
2. a. P1P2 = a cm;P2P3 =
cm;P3P4 = a cm
b. Since QP1R is an isosceles right triangle, and all the cuts are perpendiculars, then P1P2 = 2P3P4 = 4P5P6 = . So, this
geometric sequence has common ratio of . Also, P2P3 = 2P4P5 = 4P6P7 = . Finally, we have P1P2 = P2P3 . So, we have P1P2 =
P2P3 = 2 P3P4 = 2 P4P5 = . So, this is a geometric sequence with common ratio 1/ or
c.a(2 + ) cm
Take the Challenge
1. a.
cm
2
b.2 +
cm
c. s = or
2. a. OP2 = 5 cm; A1 = 25 cm2
b. A2 =
cm
2
c. i. ratio =
ii. 50 cm2
iii. We know that since OPnPn + 1 is an isosceles triangle, we know that OPn + 1 = PnPn + 1 =
OPn. Also,
A1 = (OPn + 1) (PnPn + 1) = (
OPn
)
2 = ()(OPn
2) = (ln)
2
Also, A1 + A2 + = 4(l12 + l2
2 + ...) = 50
l12 + l2
2 + ...= 50/4 = 25/2
3. 15/8 mm
4. a. 40 cm
b. 40 cm
c. 400/3 cm2
Section 1.5
A.
1. 5/23, 5/26, 5/29 2. 1/11, 1/13, 1/15
3. 1/22, 1/27, 1/32 4. 1/27, 1/31, 1/35
5. 3/7, 3/8, 1/3 6. 4/23, 1/7, 4/33
B.
1. 1/ 79 2. 2/23
3. 1/18 4. Not a harmonic sequence
-
Powerplus
A.
1. Impossible. The terms of an arithmetic sequence are 2/5, 1/5, 0, - 1/5, - 2/5. However, taking the reciprocals of each will
result in one having an undefined value.
2. 12/11, 12/7, 4, - 12
3. 12/25, 6/19
4. 1/5, 1/8, 1/11
5. 1/10; 2/n
6. , 1/3, , 1/5, .
7. 9 and 1
8. 2
9. 6 and 24
10. It is given
Take the Challenge
1. First, since m and n are in a harmonic progression, then 1/m and 1/n are in an arithmetic progression. Let a be the first term
in that sequence. So,
1/m = a + (n 1)d => 1 = am + dm(n 1), and
1/n = a + ( m 1)d => 1 = an + dn(m 1). Subtracting the second from the first, we get
1 1 = am an + dm(n 1) dn(m 1) => 0 = a(m n) + dmn dm dmn + dn => 0 = a(m n) d(m n) => 0 = (a d)(m n).
We know that m n, so, we have a d = 0 or a = d. Substituting to any of the equations (both will yield the same result, but for
the sake of this solution lets use the 1st
equation.)
1/m = a + (n 1)d = a + (n 1)a = an => a = 1/mn.
Now, using the formula to get the (m + n)th term, we have a + (m + n 1)d = a + (m + n 1)a = a(m + n) =
So, this is the
(m + n)th term of the arithmetic progression so that its reciprocal is the harmonic progression given. Therefore, the (m + n)th
term of the harmonic progression is
.
2. Since a, b and c are terms of a harmonic progression, 1/a, 1/b and 1/c are corresponding terms of an arithmetic progression.
Let such an arithmetic progression have a general formula of a + (k 1)d. So,
1/a = a + (p 1)d.(i)
1/b = a + (q 1)d (ii)
1/c = a + (r 1)d..(iii)
-
(i) (ii) gives 1/a 1/b = a + pd d a qd + d = (p q)d => (b a) = (p q)d x ab(iv)
(ii) (iii) gives 1/b 1/c = a + qd d a rd + d = (q r)d => (c b) = (q r)d x bc(v)
(iii) (i) gives 1/c 1/a = a + rd d a pr + d = (r p)d => (a c) = (r p)d x ac.(vi)
Adding (iv), (v), and (vi) yields
(b a) + (c b) + (a c) = (p q)d x ab + (q r)d x bc + (r p)d x ac
0 = d[(p q)ab + (q r)bc + (r p)ac]. But d 0, so
bc(q r) + ca(r p) + ab(p q) = 0, which gives out the desired result.
3. 6 and 3
4.
5.
Section 1.6
A.
1. 6 2. 120 3. 2 4. 1
5. 45 6. 84 7. 7 8. 28
9. 20 10. 11 11. 1 12. 105
B.
13. c5 + 10c
4 + 40c
3 + 40c
2 + 80c + 32 14. 81+108a +54a
2 +12a
3 +a
4
15. x6 + 6x
5y + 15x
4y
2 + 20x
3y
3 + 15x
2y
4 + 6xy
5 + y
6 16. x
6 + 12x
5y + 60x
4y
2 + 160x
3y
3 + 240x
2y
4 + 192xy
5 + 64y
6
17. x5
+ 15x4y + 90x
3y
2 +270x
2y
3 + 405xy
4 +243y
5 18.
19. y16
- 4y12
+ 6y8 4y
4 + 1 20. 32x
4 + 96x
3y + 216x
2y
2 + 216xy
3 + 81y
4
21. x8 + 8x
6y + 24x
4y
2 + 32x
2y
3 + 16y
4 22. 8x
6 12x
4y
2 + 6x
2y
4 y
6
23. 27x6 + 54x
4y
4 + 36x
2y
8 + 8y
12 24.
25.
26. x
6 - 6 x
5 + 30x
4 - 40 x
3 + 60x
2 - 24 x + 8
27. y6 - 6 y
5 + 45y
4 - 60 y
3 + 135y
2 - 54 y + 27 28. 8a
3/2 + a
2 + 24a + 32a
1/2 + 16
29. a4/3
+ 24a2/3
+ 8a + 32a1/3
+ 16 30.
C.
31. x16
+ 8x15
+ 28x14
32. a28
+ 14a27
+ 91x26
33. 256a8 + 1024a
7b + 1792a
6b
2 34. a
12 + 36a
11 + 594a
10
-
35. a8 16a
7b + 112a
6b
2 36. 128a
7 2240a
6 + 16800a
5 37. a
20 + 10a
18b
2 + 45a
16b
4 38. a
28 14a
26b
2 + 91a
24b
4
39. a42
+ 42a41
b + 861a
40b
2 40. a
93 93a
92b + 4278a
91b
2 41. y
7 + 7y
5 + 21y
3 42. y
8 +
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