Pertanyaan
dari sebuah deret aritmatika diketahui U3=9 sedangkan jumlah U5 dan U7 sama dengan 36 tentukan jumlah 10suku pertama deret tersebut
Ditanyakan oleh: USER4961
131 Dilihat
131 Jawaban
Jawaban (131)
U3 = a + 2b = 9
U5 + U7 = 36
a + 4b + a + 6b = 36
2a + 10b = 36
2a + 10b = 36 |x1| 2a + 10b = 36
a + 2b = 9 |x2| 2a + 4b = 18
6b = 18
b = 18/6 = 3
a + 2b = 9
a + 2(3) = 9
a = 9-6 = 3
S10 = 10/2 (2(3) + 9(3))
= 5 (6+27)
= 5(33)
= 165
U5 + U7 = 36
a + 4b + a + 6b = 36
2a + 10b = 36
2a + 10b = 36 |x1| 2a + 10b = 36
a + 2b = 9 |x2| 2a + 4b = 18
6b = 18
b = 18/6 = 3
a + 2b = 9
a + 2(3) = 9
a = 9-6 = 3
S10 = 10/2 (2(3) + 9(3))
= 5 (6+27)
= 5(33)
= 165
u3 = a + (n-1)b
9 = a+2b....persamaan 1 (kalikan 2)
a+4b + a+6b = 36
2a + 10b = 36..... persamaan 2
eliminasikan
2a +4b = 18
2a + 10b = 36
-6b = -18
b = 3
a + 2b = 9
a + 6 = 9
a = 3
u10 = 3+ 9×3 = 30
S10= n/2 ( a+un)
= 10/2( 3+30)
= 5×33 = 165
9 = a+2b....persamaan 1 (kalikan 2)
a+4b + a+6b = 36
2a + 10b = 36..... persamaan 2
eliminasikan
2a +4b = 18
2a + 10b = 36
-6b = -18
b = 3
a + 2b = 9
a + 6 = 9
a = 3
u10 = 3+ 9×3 = 30
S10= n/2 ( a+un)
= 10/2( 3+30)
= 5×33 = 165