Question

Dada uma PA de 4 termos onde a soma dos dois primeiros termos é - 9 e a soma dos dois últimos é 19.O produto entre o segundo e o terceiro termo é igual a:

a) - 6
b) - 1
c) 5
d) 8​

Answer 1

$\large\boxed{\begin{array}{l}\underline{\sf Notac_{\!\!,}\tilde ao\,especial\,para\,PA\,de\,4\,termos}\\\rm (x-3r,x-r,x+r,x+3r)\end{array}}$

$\large\boxed{\begin{array}{l}\rm (x-3r,x-r,x+r,x+3r)\\\rm a_1+a_2=-9\\\rm x-3r+x-r=-9\\\rm 2x-4r=-9\\\rm a_3+a_4=19\\\rm x+r+x+3r=19\\\rm 2x+4r=19\\\sf montando\,um\,sistema\,temos:\\\begin{cases}\rm 2x-4r=-9\\\rm2x+4r=19\end{cases}\end{array}}$

$\large\boxed{\begin{array}{l}\sf Aplicando\,o\,m\acute etodo\,da\,adic_{\!\!,}\tilde ao,somemos\\\sf as\,equac_{\!\!,}\tilde oes\,membro\,a\,membro.\\+\underline{\begin{cases}\rm2x-\diagdown\!\!\!\!\!\!4r=-9\\\rm2x+\diagdown\!\!\!\!\!4r=19\end{cases}}\\\rm 4x=10\\\rm x=\dfrac{10\div2}{4\div2}=\dfrac{5}{2}\\\\\rm 2x+4r=19\\\rm \backslash\!\!\!2\cdot\dfrac{5}{\backslash\!\!\!2}+4r=19\\\rm 5+4r=19\\\rm 4r=19-5\\\rm 4r=14\\\rm r=\dfrac{14\div2}{4\div2}=\dfrac{7}{2}\end{array}}$

$\large\boxed{\begin{array}{l}\rm a_2\cdot a_3=(x-r)\cdot (x+r)\\\rm a_2\cdot a_3=x^2-r^2\\\rm a_2\cdot a_3=\bigg(\dfrac{5}{2}\bigg)^2-\bigg(\dfrac{7}{2}\bigg)^2\\\\\rm a_2\cdot a_3=\dfrac{25}{4}-\dfrac{49}{4}\\\\\rm a_2\cdot a_3=-\dfrac{24}{4}\\\\\rm a_2\cdot a_3=-=6\\\huge\boxed{\boxed{\boxed{\boxed{\rm\dagger\red{\maltese}~\blue{alternativa~a}}}}}\end{array}}$