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Mamedov12
@Mamedov12
August 2022
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Запишите семь начальных членов последовательности с общим членом а "n":
a) a"n"= n+3
b) a"n"= 1-0,1n
c) a"n"= 3+1/3n
d) a"n"= 3n-5
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Evgenia4836
Verified answer
А) a1=1+3=4
a2=2+3=5
a3=3+3=6
a4=4+3=7
a5=5+3=8
a6=6+3=9
a7=7+3=10
б) a1= 1-0,1= 0,9
a2=1-0,2= 0,8
a3= 1-0,3= 0,7
a4= 1-0,4= 0,6
a5= 1-0,5=0,5
a6= 1-0,6= 0,4
a7=1-0,7= 0,3
в) a1= 3+1/3= 3 1/3
a2= 3+ 2/3= 3 2/3
a3= 3+3/3= 4
a4= 3+ 4/3= 4 1/3
a5= 3+ 5/3= 4 2/3
a6= 3+ 6/3= 5
a7= 3+ 7/3= 5 1/3
г) a1= 3-5= -2
a2= 3*2-5= 1
a3= 3*3-5= 4
a4= 3*4-5= 7
a5= 3*5-5= 10
a6= 3*6-5= 13
a7= 3*7-5= 16
2 votes
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Answers & Comments
Verified answer
А) a1=1+3=4a2=2+3=5
a3=3+3=6
a4=4+3=7
a5=5+3=8
a6=6+3=9
a7=7+3=10
б) a1= 1-0,1= 0,9
a2=1-0,2= 0,8
a3= 1-0,3= 0,7
a4= 1-0,4= 0,6
a5= 1-0,5=0,5
a6= 1-0,6= 0,4
a7=1-0,7= 0,3
в) a1= 3+1/3= 3 1/3
a2= 3+ 2/3= 3 2/3
a3= 3+3/3= 4
a4= 3+ 4/3= 4 1/3
a5= 3+ 5/3= 4 2/3
a6= 3+ 6/3= 5
a7= 3+ 7/3= 5 1/3
г) a1= 3-5= -2
a2= 3*2-5= 1
a3= 3*3-5= 4
a4= 3*4-5= 7
a5= 3*5-5= 10
a6= 3*6-5= 13
a7= 3*7-5= 16