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34-22 = 12 = 3
11-7 4
cari suku awal
U7=a+(7-1)b=a+6b
22 = a + ( 6 x 3)
22 = a+ 18
a=22-18 = 4
maka a=4, b=3
Sn=n/2(2a+(n-)b)
S18= 18/2 (2.4 + (18-1) 3)
S18 = 9 (8+(17x3)
S18 = 9 (8+51)
S18 = 9 (59) = 531
U₇ = 22
22 = a + 6b ... ( pers. 1 )
U₁₁ = 34
34 = a + 10b ... ( pers. 2 )
Eliminasikan kedua persamaan :
a + 6b = 22
a + 10b = 34 -
-4b = -12
b = 3
Substitusikan nilai b = 3 ke salah satu persamaan :
a + 6b = 22
a + 18 = 22
a = 4
Maka, jumlah 18 suku pertama :
Sn = 1/2 × n ( a + ( a + ( n - 1 ) b ) )
S₁₈ = 1/2 × 18 ( 4 + ( 4 + ( 18 - 1 ) 3 ) )
S₁₈ = 9 ( 4 + ( 4 + 17 × 3 ) )
S₁₈ = 9 ( 4 + 55 )
S₁₈ = 9 × 59
S₁₈ = 531