Ile poczatkowych wyrazow ciagu artymetycznego 2,5,8,... nalezy dodac,aby otrzymac liczbe 345?

[latex]345= frac{2+2+3(n-1)}{2} cdot n \ 690= (2+2+3(n-1)) cdot n \ 690= (4+3n-3) cdot n \ 690= (3n+1) cdot n \ 690=3n^2+n \ 3n^2+n-690=0 \ Delta=1^2-4 cdot 3 cdot (-690)=1+8280=8281 \ sqrt{Delta}=91 \ n_1=frac{-1-91}{2 cdot 3} extless 0 otin N_+ \ n_2=frac{-1+91}{2 cdot 3}=frac{90}{6}=15 in N_+ [/latex] Pietnascie.

[latex]a_1=2[/latex] [latex]a_2=5[/latex] [latex]r=a_2-a_1=5-2=3[/latex] [latex]a_n=a_1+r(n-1)[/latex] [latex]a_n=2+3(n-1)=2+3n-3=3n-1[/latex] [latex]S_n=frac{a_1+a_n}{2} cdot n[/latex] [latex]frac{2+3n-1}{2} cdot n=345[/latex] [latex]frac{3n+1}{2} cdot n=345[/latex] [latex]n(3n+1)=690[/latex] [latex]3n^2+n-690=0[/latex] [latex]Delta=1-4 cdot 3 cdot (-690)=1+8280=8281[/latex] [latex]sqrt{Delta}=91[/latex] [latex]n=frac{-1-91}{6}=-15frac{1}{3} vee n=frac{-1+91}{6}=15[/latex] Odp. 15.