a.S10
b.S15
c.S20
2).diketahui deret aritmatika 2+4+6+8 tentukan
a. Jumlah suku 10 pertama
b. Suku ke 8
Penjelasan dengan langkah-langkah:
- 1). suku deret aritmatika 3+8+13+18
suku pertama (a) = 3.
beda (b) = U2 - U1 = 8 - 3 = 5.
beda (b) = U4 - U3 = 18 - 13 = 5.
a). S10
Sn = ½n (2a + (n – 1)b
S10 = ½10 (2(3) + (10 – 1)5
S10 = 10 × (6 + 9(5))/2
S10 = 10 × (6 + 45)/2
S10 = 10 × 51/2
S10 = 510/2
S10 = 255
b). S15
Sn = ½n (2a + (n – 1)b
S15 = ½15 (2(3) + (15 – 1)5
S15 = 15 × (6 + 14(5))/2
S15 = 15 × (6 + 70)/2
S15 = 15 × 76/2
S15 = 1.140/2
S15 = 570
c). S20
Sn = ½n (2a + (n – 1)b
S20 = ½20 (2(3) + (20 – 1)5
S20 = 20 × (6 + 19(5))/2
S20 = 20 × (6 + 95)/2
S20 = 20 × 101/2
S20 = 2.020/2
S20 = 1.010
- 2). diketahui deret aritmatika 2+4+6+8
suku pertama (a) = 2.
beda (b) = U2 - U1 = 4 - 2 = 2.
beda (b) = U4 - U3 = 8 - 6 = 2.
a). Jumlah suku 10 pertama
Sn = ½n (2a + (n – 1)b
S10 = ½10 (2(2) + (10 – 1)2
S10 = 10 × (4 + 9(2))/2
S10 = 10 × (4 + 18)/2
S10 = 10 × 22/2
S10 = 220/2
S10 = 110
b). suku ke-8
Un = a + (n - 1)b
U8 = 2 + (8 - 1)2
U8 = 2 + 7(2)
U8 = 2 + 14
U8 = 16
~℘₊˚✧ ﾟ
[answer.2.content]
