Question

Determine a soma dos 60 primeiros termos da P.A em que; 2a1 + a3 = - 11 e a2 + a23 = 5.
por favor. como se faz esse exercicio?

Answer 1

$\boxed{a_n=a_1+(n-1)r}$ - importante !

$\left \{ {{2a_1+a_3=-11} \atop {a_2+a_{23}=5}} \right. \\ a_n=a_1+(n-1)r\\ a_3=a_1+(3-1)r\ \to \ a_3=a_1+2r\\ a_2=a_1+(2-1)r\ \to \ a_2=a_1+1r\ \to \ a_2=a_1+r\\ a_{23}=a_1+(23-1)r\ \to \ a_{23}=a_1+22r\\ \left \{ {{2a_1+\underbrace{a_1+2r}_{a_3}=-11} \atop {\underbrace{a_1+r}_{a_2}+\underbrace{a_1+22r}_{a_{23}}}=5} \right. \\ \left \{ {{3a_1+2r=-11\ \ \ \ \ |\cdot(-2)} \atop {2a_1+23r=5}\ \ \ \ \ \ |\cdot3} \right. \\ _{_+} \left \{ {{-6a_1-4r=22} \atop {6a_1+69r=15}} \right. \\ -------$
$65r=37\ \ \ \ \ |:65\\ r=\frac{37}{65}\\ 3a_1+2r=-11\ \to \ 3a_1+2\cdot\frac{37}{65}=-11\\ 3a_1+\frac{74}{65}=-11\\ 3a_1=-11-\frac{74}{65}\\ 3a_1=-\frac{715}{65}-\frac{74}{65}\\ 3a_1=-\frac{789}{65}\ \ \ \ \ |:3\\ a_1=-\frac{789}{195}\\ a_n=a_1+(n-1)r\\ a_n=-\frac{789}{195}+(n-1)\cdot\frac{37}{65}\\ a_n=-\frac{789}{195}+\frac{37}{65}r-\frac{37}{65}\\ a_n=\frac{37}{65}r-\frac{789}{195}-\frac{111}{195}\\ a_n=\frac{37}{65}n-\frac{900}{195}\\ a_n=\frac{37}{65}n-\frac{60}{13}$
$\boxed{S_n=\frac{a_1+a_n}2\cdot n}; \ \ \ n=60, \ a_1=-\frac{789}{195}, \\ a_{60}=\frac{37}{65}\cdot 60-\frac{60}{13}=\frac{2220}{65}-\frac{60}{13}=\frac{6660}{195}-\frac{900}{195}=\frac{5760}{195} \\ S_{60}=?\\ S_{60}=\frac{a_1+a_{60}}2\cdot60\\ S_{60}=(a_1+a_{60})\cdot30\\ S_{60}=(-\frac{789}{195}+\frac{5760}{195})\cdot30=\frac{4971}{195}\cdot30=\frac{4971}{13}\cdot2=\frac{9942}{13}\\ \boxed{S_{60}=\frac{9942}{13}=764\frac{10}{13}}$