Numerele 1,7,13 sunt termenii unei progresii aritmetice in care a1=1 ,an=x si ratia 7-1=6
Suma unei progresii aritmetice
Sn=(a1+an)*n/2=[a1+a1+(n-1)*r]*n/2
Facem inlocuirile
280=(1+1+(n-1)*6)*n/2 =>
560=(2+6n-6)*n
560=6n²-4n=>
3n²-2n-280=0 rezovi ecuatia
Δ=1+3*280=841
n1=[1+√841]/3=(1+29)/3=10
n2=(1-29)/3 =-28/3∉N
deci n=10
x=an=1+(10-1)*6=1+54=55
