Question

Quantos termos tem a PA (1, 4, 7, ...) e cuja soma é 145?​

Answer 1

$\boxed{\begin{array}{l}\underline{\rm soma~dos~termos~de~uma~PA}\\\sf S_n=\dfrac{n\cdot (a_1+a_n)}{2}\\\sf 145=\dfrac{n\cdot(1+a_n)}{2}\\\sf n\cdot(1+a_n)=290\\\sf a_n=a_1+(n-1)\cdot r\\\sf a_n=1+(n-1)\cdot3\\\sf a_n=1+3n-3\\\sf a_n=3n-2\\\sf n\cdot(1+3n-2)=290\\\sf n\cdot(3n-1)=290\\\sf 3n^2-n-290=0\end{array}}$

$\boxed{\begin{array}{l}\sf 3n^2-n-290=0\\\sf\Delta=b^2-4ac\\\sf\Delta=(-1)^2-4\cdot3\cdot(-290)\\\sf \Delta=1+3480\\\sf\Delta=3481\\\sf n=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\sf n=\dfrac{-(-1)\pm\sqrt{3481}}{2\cdot3}\\\sf n=\dfrac{1\pm59}{6}\\\sf n=\dfrac{1+59}{6}=\dfrac{60}{6}=10~termos\end{array}}$