Grade 12 - More Exercises.

Linear number patterns,arithmetical sequences : answers.

1.
Consider the following pattern / arithmetical sequence : 8; 13; 18; 23; . . .
1.1
Common difference d = T2 - T1 = 13 - 8 = 5 The next two terms are 23+5 = 28 and 28+5 = 33
1.2
a = T1 = 8 and d = 5 so that Tn = a + (n-1)d = 8 + (n - 1) X 5 = 5n + 3
1.3
T18 = 5(18) + 3 = 93
1.4
Tn = 108 so that 5n + 3 = 108 and thus n = 21 [ Remember Tn = 5n + 3 in 1.2 ]
1.5
Tn > 92 so that 5n + 3 > 92 and thus n > 89 5 = 17,8 so that n = 18
2.
Consider the following pattern / arithmetical sequence : 21; 27; 33; 39; . . .
2.1
Common difference d = T2 - T1 = 27 - 21 = 6 The next two terms are 21+6 = 27 and 27+6 = 33
2.2
a = T1 = 21 and d = 6 so that Tn = a + (n-1)d = 21 + (n - 1) X 6 = 6n + 15
2.3
T21 = 6(21) + 15 = 141
2.4
Tn = 81 so that 6n + 15 = 81 and thus n = 11
2.5
Tn > 200 so that 6n + 15 > 200 or n = 30,8333 so that n = 31 [Tn = 201]
3.
Consider the following pattern / arithmetical sequence : 62; 57; 52; 47; . . .
3.1
Common difference d = T2 - T1 = 57 − 62 = −5
The next two terms are 47 + (−5) = 42 and 42 + (−5) = 37, thus 42 and 37
3.2
a = T1 = 62 and d = −5
3.3
T8 = 67 −5(8)
Tn = a + (n-1)d = 62 + (n - 1) X (−5)
= 27
= 67 − 5n
3.4
Tn = −38   :    67 − 5n = −38
3.5
Tn > 0  :   67 − 5n > 0
−5n = −67  − 38
5n < 67 and thus n < 13,4
n = 21
n = 14
3.6
Tn < 0  :   67 − 5n > 67    n = 14
4.
Consider the following pattern / arithmetical sequence : 36; 30; 24; 18; . . .
4.1
Common difference d = T2 - T1 = 30 − 36 = −6
The next two terms are 18 + (−6) = 12 and 12 + (−6) = 6, thus 12 and 6
4.2
a = T1 = 36 and d = −6    Tn = 36 + (n - 1) X (−6) = 42  − 6n
4.3
T15 =  42 − 6(15)
4.4
T 15  =  − 24   :   42  − 6n = − 24
= − 48
6n = 42 + 24
n = 11
4.5
Tn = 0   :   42  − 6n = 0
4.6
Tn = 0   :   42  − 6n < 0
6n = 42
6n > 42   so that n > 7
n = 7
n = 8
5.
Consider the following pattern / arithmetical sequence : - 68; - 61; - 54; - 47; . . .
5.1
Common difference d = T2 - T1 =  −61 −(−68) =  7   and a = −68
The next two terms are  −47 + 7 =  −40  and  −40 + 7 = −33, thus  −40 and −33
5.2
Tn = a + (n − 1)d
5.3
T31 = 7(31) − 75
Tn = −68 + (n − 1)(7)
= 217 − 75
= 7n −75
= 142
5.4
Tn = −19 :  7n − 75 = −19
5.5
Tn > 0 :  7n − 75 > 0
7n = 56
7n > 75    i.e. n > 10,714.....
n = 8
n = 11
6.
Consider the following pattern / arithmetical sequence : - 8; - 12; - 16; - 20; . . .
6.1
d = T2 - T1 =  −12 −(−8) =  −4
6.2
Tn = a + (n − 1)d
The next two terms are −20 + −4 = −24
Tn = (−8) + (n − 1)(−4)
and −24 + −4 = −28
= −4n −4
6.3
T15 = −4(15) −4
6.4
T n = −19  :  −4n −4 = −19
= −60 −4
4n = 40−4  = 36
= −64
n = 9
6.5
Tn < − 99 :  −4n −4 < − 99
6.6
T23 - T21 = [−4(23) −4] − [−4(21) −4]
4n > 95   i.e. n > 23,75
= ( −92 −4) − (−84 −4)
n = 23
= (−96) − (−88) = −8
7.1
T1 = 7n - 3 :  Tn = 7(1) - 3
7.2
T22 - T18 = [7(22) - 3] − [7(18) - 3]
= 4
= [154 - 3] − [126 - 3]
T2 = 7n - 3 :  T n = 7(2) - 3
= 151 − 123
= 6
= 28
8.1
T21 = 89  :    a + (21-1)d = 89
8.2
a = 9 and d = 4
a + 20d = 89 . . . (1)
Tn =  a + (n-1)d
T33 =137  :    a + (33-1)d = 137
=  9 + (n-1)(4)
a + 32d = 137 . . . (2)
=  4n + 5
(2) - (1) : 
12d = 48
d = 4
8.3
T26 = 4n + 5
In (1) : 
a + 20(4) = 89
T26 = 4(26) + 5
a = 9
= 109
T1 = 9
T2 = 9  + 4 = 13
9.1
T11 =  − 11  :    a + (11-1)d = − 11
9.2
a = 19 and d = − 3
a + 10d = − 11 . . . (1)
Tn > 0  :    a + (n-1)d > 0
T25 = − 53  :    a + (25-1)d = − 53
19 + (n-1)(− 3) > 0
a + 24d = − 53 . . . (2)
19 − 3n + 3 > 0
(2) - (1) : 
14d = − 42
22 > 3n
d = − 3
n < 7,333.....
In (1) : 
a + 10(− 3) = − 11
n  = 7
a = 19
T17 = 19 + (17 - 1)(− 3)
= − 29
10.1
T21 =  73  :    a + (21-1)d = 73
10.2
a =  − 67 and d = 7
a + 20d = 73 . . . (1)
Tn > 0  :    a + (n-1)d > 0
T43 = 227  :    a + (43-1)d = 227
− 67 + (n-1)(7) > 0
a + 42d = 227 . . . (2)
− 67 + 7n − 7 > 0
(2) - (1) : 
22d = 154
7n > 74
d = 7
n < 10,57.....
In (1) : 
a + 140 = 73
n  = 11
a = − 67
T18 = − 67 + (18 - 1)(7)
= 52
11.1
T1 =  13 = a
11.2
a =  13  and d  = − 4
T2 =  9  :   d = 9 − 13
Tn  − 15  :    a + (n-1)d =  − 15
T12 =  a + (12-1)d
13 + (n-1)(− 4) = − 15
= 13 + 11(−4)
13 − 4n + 4 = − 15
= 13 − 44
− 4n = − 32
= − 31
n = 8
12.1
Determine the value of d for the sequence.
12.2
a = − 4  and d = (−5) − (−4) = − 1
d =  T2 - T1 = T3 - T2
T9 = a + (n − 1)d
(5x) − (3x − 1) = (8x − 2) − (5x)
= − 4 + (9 − 1)(− 1)
3 = x
= − 4 − 9 + 1
T1 = 3(3) − 1 = 8
= −12
T2 = 5(3) = 15
T3 = 8(3) − 2 = 22
13.1
d =  T2 - T1 = T3 - T2
13.2
a = 8  and d = (2) − (8) = − 6
(3x − 13) − (x + 3) = (x − 9) − (3x − 13)
T11 = a + (n − 1)d
2x − 16 = −2x + 4
= 8 + (11 − 1)(− 6)
x = 5
= 8 − 60
T1 = x + 3 = (5) + 3 = 8
= −52
T2 = 3x − 13 = 3(5) − 13 = 2
T3 = x − 9 = (5) − 9 = −4
14.1
d =  T2 - T1 = T3 - T2
14.2
Tn = 5n + 13
x − 18 = 28 − x
T8 = 5(8) + 13
2x = 46
= 40 + 13
x = 23
= 53
a = 18  and d = 23 − 18 = 5
14.3
Tn > 61 :    5n + 13 > 61
Tn = 18 + (n − 1)(5)
5n > 48
= 18 + 5n − 5
n > 9,6
= 5n + 13
n = 10
  
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