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Lina016
@Lina016
March 2022
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Дана арифметическая прогрессия (an) : 13; 0; -13....Найдите сумму первых двенадцати членов прогрессии
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Lesben
A1=13,a2=0,a3=-13,s12=?
d=a2-a1=a3-a2=......
d=0-13=-13
a12=a1+11.d, a12=13+11.(-13)=13-143=-130, a12=-130
s12=12/2(a1+a12)=6(13-130)=6.(-117)=-702
s12=-702
1 votes
Thanks 0
lapo4ka2803
S
=(2a1+(n-1)d)*n\2=2*13+11(-13)*12\2=(26-143)*12\2=702
0 votes
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lapo4ka2803
-703
lapo4ka2803
тьфу -702
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Answers & Comments
d=a2-a1=a3-a2=......
d=0-13=-13
a12=a1+11.d, a12=13+11.(-13)=13-143=-130, a12=-130
s12=12/2(a1+a12)=6(13-130)=6.(-117)=-702
s12=-702