Question

Sa se arate ca numerele 9,10,11 nu pot fi termeni ai unei progresii geometrice.

Answer 1

$\displaystyle Sa~presupunem~ca~ar~exista~o~p.g.~(b_n)_{n \in \mathbb{N}}} ~si~numerele \\\\ naturale~m,n,p~astfel~incat~b_m=9,~b_n=10,~b_p=11. \\\\ Notand~b_0=b~si~q=ratia,~vom~avea: \\ \\ bq^m=9 \\\\ bq^n=10 \\\\ bq^p=11 \\\\ Impartim~primele~doua~relatii~si~ultimele~doua~relatii.\\\\ Obtinem~q^{m-n}= \frac{9}{10}~(*)~si~q^{n-p}= \frac{10}{11}~(**).\\\\ O~ridicam~pe~prima~la~n-p,~iar~pe~a~doua~la~m-n. \\\\ Rezulta~q^{(m-n)(n-p)}= \left( \frac{9}{10} \right)^{n-p}= \left( \frac{10}{11} \right)^{m-n}.$

$\displaystyle Rezulta~9^{n-p} \cdot 11^{m-n}=10^{m-p}. (***) \\ \\ Trecem~la~modul~in~relatiile~(*)~si~(**). \\ \\ (Fac~asta~pentru~a~trata~cazurile~q<0~si~q>0~deodata.) \\ \\ |q|^{m-n}= \frac{9}{10}~si~|q|^{n-p}=\frac{10}{11}. \\ \\ \bullet Daca~|q| \in (0,1),~atunci~m-n>0~si~n-p>0.~Deci~m>n>p. \\ \\ Rezulta~m-p>0. \\ \\ Deci~in~(***)~toti~exponentii~sunt~naturali~nenuli.~Rezulta~ca \\ \\ MS~este~divizibil~cu~11,~pe~cand~MD~nu~este.$

$\bullet~ Daca~|q|>1,~atunci~m-n<0~si~n-p<0.~Deci~m<n<p. \\ \\ Deci~m<n<p.~Rezulta~m-p<0. \\ \\ Atunci~n-m,~p-n,~p-m~sunt~naturale~nenule. \\ \\ Ridicand~relatia~(***)~la~-1,~rezulta~ \\ \\ 9^{p-n} \cdot 11^{n-m}=10^{p-m}. \\ \\ MS~este~divizibil~cu~11,~iar~MD~nu~este. \\ \\ Deci~am~ajuns~la~o~contradictie~in~ambele~cazuri,~ceea~ce \\ \\ inseamna~ca~nu~exista~p.g.~cu~proprietatea~din~enunt.$