Question

СРООООЧНО
Послідовність(bn)-геометрична прогресія у якої b5+b4=72, b4-b2=18. ЗнайдітьS8

Answer 1

Відповідь:

$\left \{ {{b_{5}+b_{4}=72 \atop {b_{4}-b_{2}=18}} \right. \\\\b_{5}=b_{4}*q^3=b_{1}*q^4\\b_{4}=b_{1}*q^3\\b_{2}=b_{1}*q\\\\\left \{ {{b_{1}*q^4+b_{1}*q^3=72} \atop {b_{1}*q^3-b_{1}*q=18}} \right. \\\\\left \{ {{bq^3(q+1)=72} \atop {bq(q^2-1)=18}} \right. \\\\\left \{ {{bq^3(q+1)=72} \atop {bq(q-1)(q+1))=18}}$

делим уравнения друг на друга

$\frac{b_{1}q(q-1)(q+1)}{b_{1}q^3(q+1)} =\frac{18}{72} \\\\\frac{b_{1}q(q-1)}{b_{1}q^3} =\frac{18}{72} \\\\\frac{q-1}{q^2}=\frac{1}{4}\\\\q^2=4q-4\\ q^2-4q+4=0\\(q-2)^2=0\\q=2\\\\\left \{ {{b_{1}q^4+b_{1}q^3=72} \atop {b_{1}q^3-b_{1}q=18}} \right. \\\\\left \{ {{b_{1}q^4+b_{1}q^3=72} \atop {q=2}} \right. \\\\\\b_{1}*2^4+b_{1}*2^3=72\\16b_{1}+8b_{1}=72\\b_{1}(16+8)=72\\\\b_{1}=\frac{72}{24}=3$

значит, b=3, q=2

$b_{8}=b_{1}*q^7=3*2^7=3*128=384$

$S_{8}=\frac{b_{1}-b_{8}*q}{1-q}=\frac{3-384*2}{1-2} =\frac{3-768}{-1}=\frac{763}{1}=763$

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