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Suma a patru numere naturale impare consecutive este1320

Răspuns :

x+x+2+x+4+x+6=1320
4x=1320-12
4x=1308
x=1308:4
x=327
numerele consecutive sunt: 327,329,331,333
|. !___!
||. !___!+2
|||. !___!+4
|V.!___!+6
|+||+|||+|V=1 320
1 320-(2+4+6)=1 308
1 308:4=327-primul nr (un segment)
327+2=329-nr doi
327+431-nr trei
327+6=433-nr patru
P:327+329+431+433=1320.