WISKUNDE
GRAAD 12
NOG OEFENINGE
Rekenkundige reekse : antwoorde.
MATHEMATICS
GRADE 12
MORE EXERCISES
Arithmetic series : answers.
1.1
a = 8; d = T3 − T2 = T2 − T1 = 5
1.2
Tn = a + (n − 1)d
T5 = 23 + 5 = 28; T6 = 28 + 5 = 33
= 8 + (n − 1)5
= 5n + 3
1.3
T18 = 5(18) + 3
1.4
Tn = 108 : 5n + 3 = 108
= 93
5n = 105
n = 21 : T21 = 108
n
1.5
Bepaal eers die formule vir Tn en vul dan
1.6
Sn = ── [2a + (n − 1)d]
2
64
die nommers van die eerste en laaste terme in.
S64 = ── [2(8) + (64 − 1)5]
2
First determine the formula for Tn and then add
= 32 (16 + 315)
the numbers of the first and last terms.
= 10 592
Tn = 5n + 3 terme van 1 tot 64 /
terms from 1 to 64
64
Σ (5n + 3)
n = 1
2.1
a = 21; d = T3 − T2 = T2 − T1 = 6
2.2
Tn = a + (n − 1)d
T5 = 39 + 6 = 45; T6 = 45 + 6 = 51
= 21 + (n − 1)6
= 6n + 15
2.3
T21 = 6(21) + 15
2.4
Tn = 81 : 6n + 15 = 81
= 141
6n = 66
n = 11 : T11 = 81
n
2.5
Tn = 6n + 15
2.6
Sn = ── [2a + (n − 1)d]
2
114
Terme van 1 tot 114 / Terms from 1 to 114
S114 = ── [2(21) + (114 − 1)6]
2
114
Σ (6n + 15)
= 57 (42 + 678)
n=1
= 41 040
3.1
a = 62; d = T3 − T2 = T2 − T1 = −5
3.2
Tn = a + (n − 1)d
T5 = 47 − 5 = 42; T6 = 42 − 5 = 37
= 62 + (n − 1)(−5)
= 67 − 5n
3.3
T8 = 67 − 5(8)
3.4
Tn = −38 : 67 − 5n = −38
= 27
5n = 67 + 38 = 105
n = 21 : T21 = −38
n
3.5
Tn = 67 − 5n
3.6
Sn = ── [2a + (n − 1)d]
2
87
Terme van 1 tot 87 / Terms from 1 to 87
S87 = ── [2(62) + (87 − 1)(−5)]
2
87
Σ (6n + 15)
= 43,5 (124 − 430)
n=1
= −13 311
4.1
a = 63; d = T3 − T2 = T2 − T1 = −6
4.2
Tn = a + (n − 1)d
T5 = 45 − 6 = 39; T6 = 39 − 6 = 33
= 63 + (n − 1)(−6)
= 69 − 6n
4.3
T23 = 69 − 6(23)
4.4
Tn = −117 : 69 − 6n = −117
= −69
6n = 69 + 117 = 186
n = 31 : T31 = −117
n
4.5
Tn = 69 − 6n
4.6
Sn = ── [2a + (n − 1)d]
2
93
Terme van 1 tot 93 / Terms from 1 to 93
S93 = ── [2(63) + (93 − 1)(−6)]
2
93
Σ (69 − 6n)
= 46,5 (126 − 372)
n=1
= −11 439
n
n
4.5
Sn = 315 : ── [2(63) + (n − 1)(−6)] = 315
4.6
Sn = 0 : ── [2(63) + (n − 1)(−6)] = 0
2
2
n[126 − 6n + 6) = 630
n[126 − 6n + 6) = 0
6n2 − 132n + 630 = 0
6n2 − 132n = 0
n2 − 22n + 105 = 0
n2 − 2n = 0
(n − 7)(n − 15) = 0
n (n − 22) = 0
n = 7 OF / OR n = 15
n = 0 OF / OR n = 22
7 terme OF 15 terme / 7 terms OR 15 terms
22 terme / 22 temrs
∴ S7 = 315 of/or S15 = 315
∴ S22 = 0
5.
Tn = a + (n − 1)d
6.
Tn = a + (n − 1)d
T1 = 3 : a = 3
T4 = 19 : a + (4 − 1)d = 19
T9 = 75 : a + (9 − 1)d = 75
a + 3d = 19 . . . (1)
3 + 8d = 75
T11 = 54 : a + (11 − 1)d = 54
8d = 72
a + 10d = 54 . . . (2)
d = 9
(2) − (1) :
7d = 35
a =3 d = 9
d = 5
n
Stel in / Into () : a + 3(5) = 19
Sn = ── [2a + (n − 1)d]
2
a = 4 d = 5
10
n
S10 = ── [2(3) + (10 − 1)(9)]
Sn = ── [2a + (n − 1)d]
2
2
21
= 5 (6 + 81)
S21 = ── [2(4) + (21 − 1)(5)]
2
21
= 425
= ── (8 + 100)
2
= 1 134
7.
Sn = 37n − 2n2
8.1
Sn = 2n2 + 6n
T8 = S8 − S7
T11 = S11 − S10
= 37(8) − 2(8)2 − (37(7) − 2(7)2)
= 2(11)2 + 6(11) − (2(10)2 + 6(10))
= 168 − 161
= 308 − 260
= 7
= 48
8.2
Sn = 2n2 + 6n : Sn = 476
2n2 + 6n − 476 = 0
n2 + 3n − 238 = 0
(n + 17)(n − 14) = 0
n = −17 OF / OR = 14
14 terme moet bymekaar getel word. /
14 terms must be added.
∴ S14 = 476
9.1
Sn = 3n2 + 4n
10.
T6 + T13 = 0
T12 = S12 − S11
a + (6 − 1)d + [a + (13 − 1)] = 0
= 3(12)2 + 4(12) − [3(11)2 + 4(11)]
a + 5d + a + 12d = 0
= 480 − 407
2a + 17d = 0 . . . (1)
8
= 73
S8 = 80 : ── [2a + (8 − 1)d] = 80
2
9.2
Sn = 480 : 3n2 + 4n = 480
4 (2a + 7d) = 80
3n2 + 4n − 480 = 0
2a + 7d = 20 . . . (2)
(3n + 40)(n − 12) = 0
(1) − (2) 10d = −20
n = −13,3 OF / OR n = 12
d = −2
12 terme / 12 terms
In / Into (1) : 2a + 17(−2) = 0
S12 = 480
a = 17
T4 = 17 + (4 − 1)(−2)
= 17 − 6
= 11
11.1
Tn = a + (n − 1)d
12.
Bereken eers die eerste drie teme om dan
a = T1 = 17
a en d te bereken. Pas dan die formule om
d = T2 − T1 = 21 − 17 = 4
die som te bereken toe.
Tn = 97 : 17 + (n − 1)(4) = 97
First calculate the first three terms to
17 + 4n − 4 = 97
calculate the values of a and d. Then apply
4n = 84
the formula to calculate the sum.
n = 21
Tn = 45 − 8n
Daar is 21 terme in die reeks. /
T1 = 45 − 8(1) = 37
The series consists of 21 terms.
T2 = 45 − 8(2) = 29
11.2
Tn = a + (n − 1)d
T3 = 45 − 8(3) = 21
Tn = 17 + (n − 1)(4)
a = 37; d = T3 − T2 = T2 − T1 = −8
= 4n + 13
n
n
11.3
Sn = ── [2a + (n − 1)d]
Sn = ── [2a + (n − 1)d]
2
2
21
25
S21 = ── [2(17) + (21 − 1)(4)]
S25 = ── [2(37) + (25 − 1)(−8)]
2
2
21
25
= ── (34 + 80)
= ── (74 − 192)
2
2
= 1 197
= − 1 475
13.
Tn = 6k + 12
14.
Tn = 3k − 11
a = T1 = 6(1) + 12 = 18
a = T1 = 3(1) − 11 = −8
T2 = 6(2) + 12 = 24
T2 = 3(2) − 11 = −5
T3 = 6(3) + 12 = 30
T3 = 3(3) − 11 = −2
d = T2 − T1 = 24 − 18 = 6
d = T2 − T1 = −5 − (−8) = 3
n
n
Sn = ── [2a + (n − 1)d]
Sn = ── [2a + (n − 1)d]
2
2
14
18
S14 = ── [2(18) + (18 − 1)(6)]
S18 = ── [2(−8) + (18 − 1)(3)]
2
2
= 7 (36 + 102)
= 9 (−16 + 51)
= 966
= 315