Question

Numa progressão geométrica,tem-se a2+a5= 84 e a3+a6= 252. Assim,o primeiro termo e a razão valem?

Answer 1

Temos que, $a_n=a_1\cdot q^{n-1}$. Assim:

$a_2=a_1\cdot q$ e $a_5=a_1\cdot q^4$.

$a_2+a_5=84~~\Rihgtarrow~~a_1\cdot q+a_1\cdot q^4=84~~\Rightarrow~~a_1\cdot q\cdot(1+q^3)=84~~(i)$.

$a_3+a_6=252~~\Rightarrow~~a_1\cdot q^2+a_1\cdot q^5=252~~\Rightarrow~~a_1\cdot q\cdot(q+q^4)=252~~(ii)$.

Fazendo $\dfrac{(ii)}{(i)}$:

$\dfrac{a_1\cdot q\cdot(q+q^4)}{a_1\cdot q\cdot(1+q^3)}=\dfrac{252}{84}$

$\dfrac{q+q^4}{1+q^3}=3$

$q^4+q=3q^3+3$

$q(q^3+1)=3\cdot(q^3+1)$.

Daí, obtemos $\boxed{q=3}$. Substituindo em $(i)$:

$a_1\cdot q\cdot(1+q^3)=84~~\Rightarrow~~a_1\cdot3\cdot(1+3^3)=84$.

$3a_1\cdot(1+27)=84~~\Rightarrow~~3a_1\cdot28=84~~\Rightarrow~~a_1=\dfrac{84}{3\cdot28}~~\Rightarrow~~\boxed{a_1=1}$.

Answer 2

Utilizando a fórmula do termo geral: $a_n=a_1\cdot q^{n-1}$, obtemos:

$a_2=a_1\cdot q^{2-1}=a_1\cdot q$

$a_5=a_1\cdot q^{5-1}=a_1\cdot q^4$

Como $a_2+a_5=84$, temos $a_1\cdot q+a_1\cdot q^4=84$.

Colocando $a_1\cdot q$ em evidência:

$a_1\cdot q\cdot(1+q^3)=84~~\Rightarrow~~a_1=\dfrac{84}{q\cdot(1+q^3)}$

Usando o mesmo raciocínio:

$a_3=a_1\cdot q^2$

$a_6=a_1\cdot q^5$

Sendo $a_3+a_6=252$, segue que, $a_1\cdot q^2+a_1\cdot q^5=252$.

Colocando $a_1\cdot q^2$ em evidência:

$a_1\cdot q^2\cdot(1+q^3)=252$

Substituindo $a_1$ por $\dfrac{84}{q\cdot(1+q^3)}$, temos:

$\dfrac{84}{q\cdot(1+q^3)}\cdot q^2\cdot(1+q^3)}=252$

"Cortando" $(1+q^3)$, segue:

$\dfrac{84}{q}\cdot q^2=252$

$84q=252~~\Rightarrow~~q=\dfrac{252}{84}~~\Rightarrow~~q=3$.

Com isso, $a_1=\dfrac{84}{q\cdot(1+q^3)}=\dfrac{84}{3\cdot(1+3^3)}=\dfrac{84}{3\cdot28}=\dfrac{84}{84}=1$.

Olha a PG:

$(1,3,9,27,81,243,\dots)$

$a_2+a_5=84~~\Rightarrow~~3+81=84$

$a_3+a_6=252~~\Rightarrow~~9+243=252$