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a fost răspuns

Sa se determine termenul de rang 20 , n+1 , n+3 al progresiei aritmetice :a.) 2,6,10,14... b.) -2,-5,-8,-11...

Răspuns :

Lucasela
termenul general = a1+(n-1)xr
2,6,10,14.........
r=6-2=4
a20=2+(20-1)x4
a20=2+19x4
a20=78
an+1=2+(n+1-1)x4
an+1=2+4n
an+3=2+(n+3-1)x4
an+3=4n+10

-2,-5,-8,-11... , r=-3
termenul general e an=-2+(n-1)(-3)
a20=-2-3x19
a20=-59
an-1=-2+(n+1-1)x(-3)
an+1=-2-3xn
an+3=-2+(n+3-1)x(-3)
an+3=-2-3xn-6
an3=-8-3xn