[latex]a_1+a_2+a_3+a_4+a_5-(a_1+a_2+a_3)=a_4+a_5=S_5-S_3[/latex] [latex]a_4+a_5=72;;;;;;i;;;;a_4-a_3=12[/latex] [latex]a_5=a_4cdot;q;;;;;i;;;;;a_3=frac{a_4}{q}[/latex] Wstaw do układu równań: [latex]a_4+a_4q=72\a_4-frac{a_4}{q}=12;/*q[/latex] [latex]a_4(1+q)=72;;;to;;;;a_4=frac{72}{1+q};;i;;q eq;1[/latex] [latex]a_4q-a_4=12q;;czyli;;a_4(q-1)=12q;;podstaw;a_4=frac{72}{1+q}[/latex] [latex]frac{72}{1+q}cdot(q-1)=12q\12q^2+12q=72q-72\12q^2-60q+72=0;/:12[/latex] [latex]q^2-5q+6=0[/latex] [latex]q=2;;;wtedy;;;;a_4=frac{72}{1+2}=24;;;czyli;a_1cdot2^3=24;;to;a_1=3\lub\q=3;;wtedy;;;a_4=frac{72}{1+3}=18;;czyli;;a_1cdot3^3=18;;to;;a_1=frac{2}{3}[/latex] W obu przypadkach liczysz [latex]S_7=[/latex] [latex]S_7=3frac{1-2^7}{1-2}=3cdotfrac{-127}{-1}=381[/latex] [latex]S_7=frac{2}{3}frac{1-3^7}{1-3}=frac{2186}{3}[/latex] Sprawdź moje rachunki...
