Question

Calcular a soma dos 40 primeiros termos da P.A. (-2;-4;-6;...).

Answer 1

$> \: resolucao \\ \\ \geqslant \: progressao \: \: aritmetica \\ \\ r = a2 - a1 \\ r = - 4 - ( - 2) \\ r = - 4 + 2 \\ r = - 2 \\ \\ = = = = = = = = = = = = = = = = = \\ \\ > \: o \: 40 \: termo \: da \: pa \\ \\ an = a1 + (n - 1)r \\ an = - 2 + (40 - 1) - 2 \\ an = - 2 + 39 \times ( - 2) \\ an = - 2 + ( - 78) \\ an = - 2 - 78 \\ an = - 80 \\ \\ = = = = = = = = = = = = = = = = = = \\ \\ > \: a \: soma \: dos \: termos \: da \: pa \\ \\ sn = \frac{(a1 + an)n}{2} \\ \\ sn = \frac{( - 2 + ( - 80) \: )40}{2} \\ \\ sn = \frac{( - 2 - 80)40}{2} \\ \\ sn = \frac{ - 82 \times 40}{2} \\ \\ sn = - 82 \times 20 \\ \\ sn = - 1640 \\ \\ \geqslant \leqslant \geqslant \leqslant \geqslant \leqslant \geqslant \leqslant \geqslant \geqslant \geqslant$

Answer 2

Resposta:

$\textsf{Segue a resposta abaixo}$

Explicação passo-a-passo:

$\mathsf{ a_n=a_1+(n-1)\cdot r }$

$\mathsf{ a_{40}= -2+ (40-1)\cdot (-2) }$

$\mathsf{ a_{40}=-2+39\cdot(-2) }$

$\mathsf{ a_{40}=-2-78}$

$\mathsf{a_{40}=-80 }$

$\mathsf{ S_n=\dfrac{(a_1+a_n)\cdot n}{2}}$

$\mathsf{ S_{40}=\dfrac{( -2-80)\cdot40}{2}}$

$\mathsf{S_{40}=\dfrac{-82\cdot 40}{2} }$

$\mathsf{S_{40}=-41\cdot40 }$

$\boxed{\boxed{\mathsf{ S_{40}= - 1640}}}$