Question

Tentukan jumlah n suku pertama pada deret-deret aritmatika jika diketahui :
a) suku ke 2 = 6 dan suku ke 11 = 24
b) jumlah 6 suku pertama = 48 dan jumlah 5 suku pertama = 37

Answer 1

A) u2 = 6, u11 = 24,
b = (u11 - u2) / ( 11 - 2) = ( 24 - 6) / 9 =18 /9 = 2

U2 = 6
a + b = 6
a + 2 = 6
a = 4

Sn = n ( 2a + ( n - 1)b) / 2
= n ( 8 + (n-1).2) / 2
= n ( 4 + (n-1)
= n( n + 3)
= n^2 + 3n

B)
S6 = 48
S5 = 37
U6 = s6 - s5 = 48 - 37 = 11

S6 = 48
6(a + u6) /2 = 48
3(a + 11)=48
3a + 33 = 48
3a = 15, a= 5

U6 = 11
a + 5b = 11
5b = 11 - a = 11 - 5 = 6
b = 6/5

Sn = n ( 2a + ( n - 1) b / 2
= n (10 + ( n-1)(6/5) /2
= n ( 5 + ( n - 1) (3/5)
= n ( 5 + 3n/5 - 3/5)
= n ( 3n/5 - 22/5)
= n ( ( 3n + 22) / 5)
= ( 3n^2 + 22n) /5