Question

Vă rog frumos!Urgent!

Answer 1

numerele pare sunt de forma 2k
2K+2K+2+2K+4+2K+6+2K+8+2K+10+2K+12=14K+42
scadem numrele impare (de forma 2k+1,dar mai mici decat 2k+12)
14K+42-(2k+1+2k+3+2k+5+2k+7+2k+9+2k+11)=
=14k+42-12k-36=>2k+6=16 =>2k=22
rezulta primul nr impar=11 rezulta ca primul nr=10
verificare 10+12+14+16+18+20+22=112
112-11-13-15-17-19-21=112-96=16
sper ca te-am ajutata!!coronita!!

Answer 2

a+c+e+g+i+k+m=>7 nr.cons.pare
b+d+f+h+j+l    =>6 nr.cons.impare (nr.aflate intre cele 7 nr.con.pare)
(a+c+e+g+i+k+m)-(b+d+f+h+j+i)=16
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a=? c=? e=? g=? i=? k=? m=?

a+a+1+a+2+a+3+a+4+a+5+a+6+a+7+a+8+a+9+a+10+a+11+a+12=16*7+16*6
13a+78=112+96
13a+78=208
13a=208-78
13a=130
a=130:13
a=10

b=10+1=11
c=10+2=12
d=10+3=13
e=10+4=14
f=10+5=15
g=10+6=16
h=10+7=17
i=10+8=18
j=10+9=19
k=10+10=20
l=10+11=21
m=10+12=22

10+12+14+16+18+20+22=16*7
112=112
11+13+15+17+19+21=16*6
96=96
112-96=16

numerele consecutive pare sunt: 10;12;14;16;18;20 si 22