Question

Determine o último termo da PA( 5, 7, 9, 11, ..., an), sabendo que a soma de seus termos é igual a 480 .

Answer 1

Equação da soma de um P.A é  $S_n=\dfrac{(a_1+a_n).n}{2}$

Sabendo que a soma da P.A é 480 e o seu primeiro termo é 5, já podemos descobrir quantos termos tem essa P.A


$480=\dfrac{(5+a_n).n}{2}\\\\\\960=(5+a_n).n\\\\\\\dfrac{960}{5+a_n}=n\\\\\\\\\mathsf{Equac\~ao~de~uma~P.A~ \'e~[a_n=a_1+(n-1).r]}\\\\\mathsf{r=a_n-a_{n-1}}\\\mathsf{r=7-5=2}$


$\mathsf{Substituindo~os~dados~na~equac\~ao:}\\\\\\a_n=5+\Big(\dfrac{960}{5+a_n}-1\Big).2\\\\\\a_n=5+\dfrac{1.920}{5+a_n}-2\\\\\\a_n-3=\dfrac{1.920}{5+a_n}\\\\\\(a_n-3)~\cdot~(5+a_n)=1.920\\\\5a_n+a_n^2-15-3a_n=1.920\\\\a_n^2+2a_n-15-1.920=0\\\\a_n^2+2a_n-1.935=0\\\\\\\Delta=b^2-4.a.c\\\Delta=2^2-4.1.(-1.935)\\\Delta=4+7.740\\\Delta=7.744$


$x=\dfrac{-b\pm\sqrt{\Delta}}{2.a}\\\\\\a_n=\dfrac{-(2)\pm\sqrt{7.744}}{2.1}\\\\\\a_n=\dfrac{-2\pm88}{2}\\\\\\a_n=\dfrac{-2+88}{2}\\\\\\a_n=\dfrac{86}{2}\\\\\\\boxed{\mathsf{a_n=43}}\\\\\\a_n'=\dfrac{-2-88}{2}\\\\\\a_n'=\dfrac{-90}{2}\\\\\\\boxed{\mathsf{a_n'=-45}}~~\mathsf{\gets N\~ao~serve~pois~a~P.A~\'e~crescente}$

Último termo da P.A é 43

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