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Sa se determine x apartine lui N din egalitatea 1+5+9+....+x=231 .

Răspuns :

[tex]a_n=1\\ r=4 \\ S_n=\frac{[2+(n-1)4]n}{2}=231\Rightarrow [1+2(n-1)]n=231\Rightarrow n=11 \\ x=1+10*4=41 [/tex]
Sn=[2a1+(n-1)r]n/2
231=[2+4(n-1)]n/2
231=n+2n^2-2n
2n^2-n-231=0
Δ=1+1848=1849
n=(1+43)/4=11
a22=a1+(n-1)r
x=1+10*4=41