Question

Дано:S2=2;S6=42.Найти знаменатель геометрической прогрессии-q.Помогите пожалуйста

Answer 1

$S_2=2$
$S_6=42$
$q-$ ?

$S_n= frac{b_1*(1-q^n)}{1-q} ,$  $q neq 1$

$left { {{S_2= frac{b_1*(1-q^2)}{1-q} } atop {{{S_6= frac{b_1*(1-q^6)}{1-q} } right.$
$left { {{ frac{b_1*(1-q^2)}{1-q} =2} atop {{{ frac{b_1*(1-q^6)}{1-q}=42 } right.$
$left { {{ frac{b_1*(1-q)(1+q)}{1-q} =2} atop {{{ frac{b_1*(1-q^3)(1+q^3)}{1-q}=42 } right.$
$left { {{b_1*(1+q)} =2} atop {{{ frac{b_1*(1-q)(1+q+q^2)(1+q^3)}{1-q}=42 } right.$
$left { {{b_1*(1+q)} =2} atop {{b_1*(1+q+q^2)(1+q^3)}=42 } right.$
$left { {{b_1*(1+q)} =2} atop {{b_1*(1+q)(1-q+q^2)(1+q+q^2)}=42 } right.$
$left { {{b_1*(1+q)} =2} atop {{2*(1-q+q^2)(1+q+q^2)}=42 } right.$
$left { {{b_1*(1+q)} =2} atop {{2*((1+q^2)^2-q^2)}=42 } right.$
$left { {{b_1*(1+q)} =2} atop {{((1+q^2)^2-q^2)}=21 } right.$
$left { {{b_1*(1+q)} =2} atop {{1+q^2+2q-q^2}=21 } right.$
$left { {{b_1*(1+q)} =2} atop {{1+2q}=21 } right.$
$1+2q}=21$
$2q}=21-1$
$2q=20$
$q=10$