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a + 6b=103
U10 =88
a + 9b=88
eliminasi a : -3 b =15
b = -5
a=133
S24 =24/2 x (2 x 133 + 23 x (-5)
S24 =1812
Jadi Jumlah 24 suku pertama adalah 1812
u10=88
s24=?
u1+6b=103
u1+9b=88 -
-3b=15
b=15:-3=-5
masukkan ke persamaan 1:
u1+6b=103
u1+6 x- 5=103
u1+(-30)=103
u1=103+30=133
sn=1/2xn(u1x2+(n-1)xb)
s24=1/2x24(133x2+(24-1)xb)
s24=12.(266+23x-5)
s24=12(266+(-115))
s24=12x111=1332