Question

A equação (5+x)+(10+x) +...+(100+x)=1110 tem como solução para x Igual a ?

Answer 1

$\Large\boxed{\begin{array}{l}\sf(5+x)+(10+x)+\dotsc+(100+x)=1110\\\sf a_1=5+x\\\sf a_2=10+x\\\sf a_n=100+x\\\sf n=?\\\sf a_n=a_1+(n-1)\cdot r\\\sf 100+x=5+x+(n-1)\cdot5\\\sf 100+x=5+x+5n-5\\\sf 5n=100-5+5+x-x\\\sf 5n=100\\\sf n=\dfrac{100}{5}\\\\\sf n=20\end{array}}$

$\Large\boxed{\begin{array}{l}\sf S_{20}=\dfrac{\diagdown\!\!\!\!\!\!20\cdot(100+x+5+x)}{\diagdown\!\!\!\!2}\\\\\sf S_{20}=10(105+2x)\\\sf 1110=10(105+2x)\\\sf 105+2x=\dfrac{111\backslash\!\!\!0}{1\backslash\!\!\!0}\\\\\sf 105+2x=111\\\sf 2x=111-105\\\sf 2x=6\\\sf x=\dfrac{6}{2}\\\\\sf x=3\checkmark\end{array}}$