Question

Sa se determine primul termen si ratia progresiei geometrice
b1+b2+b3=26
b1^2+b2^2+b3^2=364
Va rog

Answer 1

$\text{Notez ratia cu q.}\\b_1+b_2+b_3=26\\b_1+b_1\cdot q+b_1\cdot q^2=26\\b_1(1+q+q^2)=26|()^2\\b_1^2(1+q+q^2)^2=676~~~~(1)\\\\b_1^2+b_2^2+b_3^2=364\\b_1^2+b_1^2\cdot q^2+b_1^2\cdot q^4=364\\b_1^2(1+q^2+q^4)=364~~~~(2)\\\text{Impartim relatia (1) la relatia (2):}\\\dfrac{b_1^2(1+q+q^2)^2}{b_1^2(q+q^2+q^4)}= \dfrac{676}{364}\\\dfrac{(1+q+q^2)^2}{1+q^2+q^4}=\dfrac{13}{7}\\\text{Termenul de la numitor se poate descompune in }(1+q+q^2)(q^2-q+1)\\\text{Prin urmare:}$

$\dfrac{(1+q+q^2)^2}{(q^2+q+1)(q^2-q+1)}=\dfrac{13}{7}\\\dfrac{q^2+q+1}{q^2-q+1}=\dfrac{13}{7}\\7q^2+7q+7=13q^2-13q+13\\6q^2-20q+6=0\\\Delta=400-4\cdot 6\cdot 6=400-144=256\\\sqrt{\Delta}=16\\q_1=\dfrac{20+16}{12}=\dfrac{36}{12}=3\\q_2=\dfrac{20-16}{12}=\dfrac{4}{12}=\dfrac{1}{3}$

Inlocuiesti pe fiecare q in prima relatie si il afli pe b1 . Sper ca te descurci singur de aici.