06 Kunci Jawaban dan Pembahasan MAT IX

Diperoleh:
2.000 + 2.002 + 2.004 + 2.006
1.986 + 1.992 + 1.998 + 2.004
= 8.012
7.980
= 4.006
3.990
13. Jawaban: d
U8 = 44
U20 = 92
Un = a + (n + 1)b, diperoleh:
U8 = a + 7b . . . . . (i)
U20 = a + 19b . . . . . (ii)
Kurangkan (ii) dari (i) sehingga diperoleh:
a + 7b = 44
a + 19b = 92
––––––––––– –
–12a = –48
a= 4
U8 = a + 7b
⇔ 44 = 4 + 7b
⇔ 40 = 7b
⇔ b= 40
7
Sn = 1
n(2a + (n – 1)b)
2
⇔ S 21 = 1
2
× 21 (2 × 4 + (21 – 1) 40
7 )
= 21 800 21 856
(8 + ) = × = 1.284
2 7 2 7
14. Jawaban: b
Un = Sn – Sn – 1
⎛ 2
n + 3n⎞
= ⎜ ⎟
⎝ 16 ⎠ –
⎛ 2
(n − 1) + 3(n −1)
⎞
⎜ ⎟
⎝ 16 ⎠
⎛ 2
n + 3n⎞
= ⎜ ⎟
⎝ 16 ⎠ –
⎛ 2
n − 2n + 1+ 3n − 3⎞
⎜ ⎟
⎝ 16 ⎠
⎛ 2
n + 3n⎞
= ⎜ ⎟
⎝ 16 ⎠ –
2
n + n− 2
16
2 2
n − n + 3n− n+ 2
=
=
16
2n 2 +
=
16
n 1 +
8
5 + 1 6
U5 = =
8 8
7 + 1 8
U7 = =
8 8
9 + 1 10
U9 = =
8 8
U5 + U7 + U9 = 6 8 10 24
+ + = = 3
8 8 8 8
15. Jawaban: d
Un = a + (n – 1)b
⇔ U7 = a + (7 – 1)b
⇔ 17 1
= a + 6b . . . . . . . (i)
2
U19 = a + (19 – 1)b
⇔ 20 1
= a + 18b . . . . . . (ii)
2
88 Kunci Jawaban dan Pembahasan PR Matematika Kelas IX
Dari (i) dan (ii) diperoleh
17 1
= a + 6b
2
20 1
= a + 18b
2
––––––––––––– –
3 = 12b ⇔ b = 1
4
17 1
= a + 6b
2
⇔ 17 1 6
= a +
2 4
⇔ a = 17 1 6 35 3 32
– = – = = 16
2 4 2 2 2
Sn = 1
n(2a + (n – 1)b)
2
⇔ S40 = 1
1
× 40(2 × 16 + 39 ×
2 4 )
= 20(32 + 39
4 )
= 640 + 5 × 39
= 640 + 195
= 835
16. Jawaban: b
Un = a + (n – 1)b
⇔ U8 = a + (8 – 1)b
⇔ 21 1
= a + 7b . . . . . . . (i)
2
U13 = a + (13 – 1)b
⇔ 36 1
= a + 12b . . . . . . (ii)
2
Kurangkan (ii) dari (i) sehingga diperoleh:
21 1
= a + 7b
2
36 1
= a + 12b
2
––––––––––––– –
–15 = –5b
⇔ b= 3
17. Jawaban: b
Dengan sifat dasar barisan aritmetika diperoleh:
2 × suku tengah = jumlah suku tepi
⇔ 2 × (5m + 4) = (m + 6) + (24 – 2m)
⇔ 10m + 8 = m – 2m + 30
⇔ 10m + 8 = –m + 30
⇔ 11m = 22
⇔ m= 2