dalam sebuah deret aritmetika diketahui a=3,Un=73 dan u8+u11=91.tentukanlah n,sn dan s12

a = 3
U8 + U11 = 91
Un = 73

U8+U11 = 91
a+7b + a+10b = 91
2a + 17b = 91
2 (3) + 17b = 91
17 b = 91-6
17b = 85
b = 5

Un = 73
a + (n-1)b = 73
3 + (n-1)5 = 73
3 +5n -5 = 73
5n -2 =73
5n = 75
n = 15

Sn = n/2 (2a+(n-1)b)
S15 = 15/2 (2 (3)+(15-1)5)
= 15/2 (6+70)
= 15/2 (76)
= 570

S12 = 12/2 (2 (3)+(12-1)3)
= 6 (6+33)
= 6 (39)
= 234

Pelajaran : Matematika

pembahasan :
a = 3

(a + 7b) + (a + 10b) = 91
2a + 7b = 91
2(6) + 17b = 91
6 + 17b = 91
17b = 85
b = 91/17 = 5

73 = 3 + (n - 1)5
70 = 5n - 5
5n = 75
n = 75/5
S15 = 15(2(3) + (15 - 1)5) = 570
S12 = 12/2 (2(3) + 12 - 1)5) = 366

Semangat!