Jika suku pertama dalam suatu barisan aritmetika adalah -2 dengan beda 3. S(n) adalah jumlah n suku pertama deret aritmetika tersebut dan S(n+2) - S(n) = 65. Ma

diketahui :
a = -2
b = 3
s(n+2) - s(n) = 65

ditanya :
nilai n?

jawab :
s(n) = n/2{2a+(n-1)b}
s(n) = n/2{2(-2)+(n-1)3}
s(n) = n/2{(-4)+(n-1)3}
s(n) = n/2 {-4+ n3-3}
s(n) =n/2 {-7+3n}
s(n) = -7n/2 + 3n²/2

s(n+2) = n+2/2{2a+(n+2-1)b}
s(n+2) = n+2/2{2a+(n+1)b}
s(n+2) = n+2/2{2(-2)+(n+1)3}
s(n+2) = n+2/2{(-4)+ 3n+3}
s(n+2) = n+2/2{(-1)+3n}
s(n+2) = 3n²/2 +5n/2 -2/2

s(n+2) - s(n) = 65
3n²/2 +5n/2 -2/2 - [-7n/2 + 3n²/2] = 65
3n²/2 +5n/2 -2/2 + 7n/2 - 3n²/2 =65
12n/2 - 2/2 = 65 (dikalikan semua dengan 2)
12n -2 = 130
12n = 132
n= 11