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Serilpai
@Serilpai
July 2022
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В арифметической прогрессии a1+a3=12 a4=12 S5=?
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aleksandr1799
A1+a3 = 2a1+ 2d = 12 a1 + d = 6 a1 = 6-d
a4 = a1 + 3d = 12
6-d+3d = 12
2d = 6
d=3 a1 = 3
sn = a1 + d*(n-1)/2
s5= 3 + 3*(4)/2 = 9
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Answers & Comments
a4 = a1 + 3d = 12
6-d+3d = 12
2d = 6
d=3 a1 = 3
sn = a1 + d*(n-1)/2
s5= 3 + 3*(4)/2 = 9