Metoda algebrica:
a+(a+1)+(a+2 )= 81
3×a+3= 81
3×a = 81-3
3×a = 78
a = 78:3
a = 26
(a+1)= 27
(a+2) =28
Verificare:
26+27+28= 81
26;27;28 sunt cele 3 nr consecutive
Metoda grafica:
a |_____|
(a+1) |_____|+1
(a+2) |_____|+2
avem 3 segmente egale +1+2 = 81
81 -3 = 78
78:3 = 26 ( un segment)
a = 26
(a+1) = 26+1=27
(a+2) = 26+2 = 28
