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a8-a4=12
taki uklad równań
a wiadomo, że w ciagu arytmetycznym wzór na n-ty wyraz ciagu to
an= a1 + (n-1)r
r - róznica
wiec a6= a1 + 5r
i a7 = a1 + 6r
a8 = a1 + 7r
a4= a1 + 3r
podstwiamy do ukladu -
a1 + a1 + 5r + a1 + 6r = 48 czyli 3a1 + 11r= 48
i a1 + 7r - (a1 + 3r) = 12
4r = 12 r=3
3a1 + 11 razy 3 = 48
3a1 = 15
a1= 15
wzór ciagu to an = 3n + 2