Oblicz n oraz iloraz ciągu geometrycznego q wiedząc, że pierwszy wyraz wynosi1/2 an=1/128 a Sn=127/128

[latex]a_n = a_1 cdot q^{n-1}= dfrac{1}{2} cdot q^{n-1}=dfrac{1}{128} qquad /cdot 2 \ \ q^{n-1}=dfrac{2}{128}=dfrac{1}{64} \ \ \ S_n = a_1 cdot dfrac{1-q^n}{1-q}= dfrac{127}{128} \ \ \ dfrac{1}{2} cdot dfrac{1- q^{n-1} cdot q}{1-q}=dfrac{127}{128} \ \ hbox{Podstawiam pod} q^{n-1} hbox{wartosc} dfrac{1}{64}: [/latex] [latex]dfrac{1}{2} cdot dfrac{1-frac{1}{64}q}{1-q}=dfrac{127}{128} \ \ dfrac{frac{64-q}{64}}{2(1-q)}=dfrac{127}{128} \ \ \ dfrac{64-q}{128(1-q)}=dfrac{127}{128} \ \ 128(64-q)=128 cdot 127(1-q) qquad /:128 \ \ 64-q=127-127q \ \ 126q = 63 qquad /:126 \ \ q=dfrac{63}{126}=dfrac{1}{2}[/latex] A więc q wynosi 1/2. Wyznaczam teraz n: [latex]q^{n-1} = dfrac{1}{64} \ \ left(dfrac{1}{2} ight)^{n-1} = left( dfrac{1}{2} ight)^6 \ \ \ n-1=6 \ \ n=7[/latex] więc n=7.