Question

Срочно!!!
Знайти суму шести членів геометричної прогресії (bn), якщо b4-b1=52, b1+b2+b3=26

Answer 1

$\displaystyle\bf\\\left \{ {{b_{4}-b_{1} =52 } \atop {b_{1} +b_{2} +b_{3} =26}} \right. \\\\\\\left \{ {{b_{1}\cdot q^{3} -b_{1} =52 } \atop {b_{1} +b_{1}\cdot q +b_{1}\cdot q^{2} =26}} \right.\\\\\\\left \{ {{b_{1}\cdot( q^{3} -1)=52 } \atop {b_{1}\cdot(1 + q + q^{2}) =26}} \right.\\\\\\:\left \{ {{b_{1}\cdot( q -1)\cdot(1+q+q^{2}) =52 } \atop {b_{1}\cdot(1 + q + q^{2}) =26}} \right.\\----------------\\q-1=2\\\\q=3\\\\b_{1} =\frac{52}{q^{3}-1 }=\frac{52}{3^{3}-1 } =\frac{52}{26} =2$

$\displaystyle\bf\\S_{6} =\frac{b_{1}\cdot(q^{6}-1) }{q-1} =\frac{2\cdot(3^{6}-1) }{3-1}=729-1=728\\\\\\Otvet \ : \ S_{6} =728$