Oktoberlin
1. a1 = 54 ---> suku pertama barisan geometri. a5 = 2/3 a7 = ...? diketahui bahwa : a3^2 = a1 x a5 a3^2 = 54 x 2/3 a3^2 = 36 a3 = 6 kemudian, a5^2 = a3 x a7 a7= (a5^2)/a3 =((2/3)^2)/6 =(4/9)/6 =4/36
2. U1 = a = 2 -->suku pertama deret aritmatika U10 = 38 S20 = ..? --> jumlah 20 suku pertama
Sn = 1/2 x n x (2 x a+((n-1) x b)) akan dicari nilai b (beda) U10 = a+((10-1) x b)=38 2+(9 x b) = 38 9 x b = 36 b = 4 Jadi, S20 = 1/2 x 20 x (2 x 2+((20 - 1) x 4)) = 10 x (4+(19 x 4)) = 10 x (4+76) = 10 x 80 = 800
U10: A + 9B : 38 -
-9b = - 36
B = 4
sn = n / 2 ( 2a + ( 20 - 1 ) 4
20 / 2 ( 2 × 2 + 19 × 4 )
10 × 80 = 800
a5 = 2/3
a7 = ...?
diketahui bahwa :
a3^2 = a1 x a5
a3^2 = 54 x 2/3
a3^2 = 36
a3 = 6
kemudian,
a5^2 = a3 x a7
a7= (a5^2)/a3
=((2/3)^2)/6
=(4/9)/6
=4/36
2. U1 = a = 2 -->suku pertama deret aritmatika
U10 = 38
S20 = ..? --> jumlah 20 suku pertama
Sn = 1/2 x n x (2 x a+((n-1) x b))
akan dicari nilai b (beda)
U10 = a+((10-1) x b)=38
2+(9 x b) = 38
9 x b = 36
b = 4
Jadi, S20 = 1/2 x 20 x (2 x 2+((20 - 1) x 4))
= 10 x (4+(19 x 4))
= 10 x (4+76)
= 10 x 80
= 800