Question

1.diketahui deret aritmatika 3+6+9+12.. Hitunglah jumlah 16 suku pertama deret itu
2.tentukan rumus jumlah n suku pertama deret aritmatika.jika u7=15 dan sn=168

Answer 1

1.
S16 = 16/2 (2a + 15b)
S16 = 8 (2a + 15b)
S16 = 16a + 120b
S16 = 16 × 3 + 120 × 3
S16 = 3 × 136 => 408

2.
U7 = 15
Sn = 168
15 = a + 6b
168 = n/2 (2a + (n - 1)b)
15 - 6b = a
168 = n/2 (2(15-6b) + (n-1)b)
168 = n/2 (30-12b + (n-1)b)
168 × 2/n = 30-12b + (n-1)b
336/n = 30-12b + (n-1)b
336/7 = 30-12b + 6b
48 = 30 - 6b
-6b = 18
b = -3
15 = a + 6(-3)
15 = a - 18
a = 33
Cek :
168 = n/2 (2(33) + (n - 1)-3)
168 = n/2 (66 - 3n + 3)
168 = 7/2 (69 - 3(7))
168 = 7/2 (69 - 21)
168 = 7/2 (48)
Sn = n/2 (2(33) + (n-1)-3)
= n/2 (66 - 3n + 3)
= n/2 (69-3n) <=> (-3n^2 + 69n) / 2