Tentang geometri

1. PEMBAHASAN
ULANGAN DERET
GEOMETRI
TAKTERHINGGA

2. 1 B
a. -3+1+(-⅓)+…
a = -3
r = Un = U2 = 1 = - ⅓
Un-1 U1 -3
S∞ = a = -3
1-r 1-(-⅓)
= -3 = -3
1+ ⅓ 4
3
= -3 x 3 = -9
4 4 Jadi S ∞ = -9/4

3. b. 4+4/3+4/9+…
a = 4
r = Un = U2 = 4/9 = 4/9x3/4 =1/3
Un-1
U1 4/3
S∞ = a = 4
1-r 1-⅓
= 4
⅔
= 4 x 3 = 12 = 6
2 2 Jadi S ∞=6

4. a = U1 = 6
r = ⅔
S∞ = a
1-r
= 6
1- ⅔
= 6
⅓
= 6 x 3
1
= 18 Jadi S ∞ =18
2 B

5. S∞ = 3
a = 4
r = ?
S∞ = a
1-r
3 = 4
1-r
3x(1-r) = 4
(1-r) = 4/3
-r = 4/3 – 1
-r = 4/3 – 3/3
-r = 1/3
r = -1/3 Jadi r = -1/3
3 B

6. a = 60
r = 2/3
S∞ = a
1-r
= 60
1/3
= 60 x 3
1
= 180
Jadi Panjang Lintasan 180 km
4 B

7. U0 = 18
r = 2/3
U1 = 18 x 2/3
= 12 = a
S∞= U0 + 2 S∞
= 18 + 2 12
1-r(2/3)
= 18 + 2 12
1/3
= 18 + 2 12x3
1
= 18 + 2x36
= 18 + 72
= 90
5 B

8. 1A• 3+1+1/3+….
U1=a=3 r=1/3
S∞ = a = 3
1-r 1-⅓
= 3 = 3 x 3
2/3 2
= 9 = 4 1/2
2
S ∞ =4,5

9. b. 8+(-4)+2+(-1)+…
a = 8
r = -4/8 = -1/2
S∞ = a = 8
1-r 1-(-1/2)
= 4
3/2
= 8 x 2 = 16
3 3
Jadi S ∞= 16/3

10. a = U1 = 6
r = ⅔
S∞ = a
1-r
= 6
1- ⅔
= 6
⅓
= 6 x 3
1
= 18 Jadi S ∞ =18
2 A

11. a = 60
r = 2/3
S∞ = a
1-r
= 60
1/3
= 60 x 3
1
= 180
Jadi Panjang Lintasan 180 km
3 A

12. U0 = 18
r = 2/3
U1 = 18 x 2/3
= 12 = a
S∞= U0 + 2 S∞
= 18 + 2 12
1-r(2/3)
= 18 + 2 12
1/3
= 18 + 2 12x3
1
= 18 + 2x36
= 18 + 72
= 90
4 A

13. 5 Aa+1+1/a+….=4a
U1=a r=1/a
4a (1-1/a) = a
4a (a/a – 1/a) = a
4a (a-1) = a
a
4 (a-1) = a
4a-a = a
3a = 4
a = 4/3