Question

Ache a soma dos 43 primeiros termos da PA (8,2...)

Answer 1

$\sf a_n=a_1+(n-1)\cdot\underbrace{\sf a_2-a_1}_{\sf r}$

$\sf a_{43}=8+(43-1)\cdot(2-8)$

$\sf a_{43}=8+42\cdot(-6)$

$\sf a_{43}=8-252$

$\boxed{\sf a_{43}=-244}$

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

$\sf S_{43}=\dfrac{(8+(-244))\cdot43}{2}$

$\sf S_{43}=\dfrac{(8-244)\cdot43}{2}$

$\sf S_{43}=\dfrac{-236\cdot43}{2}$

$\sf S_{43}=\dfrac{-10148}{2}$

$\boxed{\boxed{\sf S_{43}=-5074}}$

Answer 2

$> resolucao \\ \\ \geqslant progressao \: aritmetica \\ \\ r = a2 - a1 \\ r = 2 - 8 \\ r = - 6 \\ \\ \\ an = a1 + (n - 1)r \\ an = 8 + (43 - 1) - 6 \\ an = 8 + 42 \times ( - 6) \\ an = 8 + ( - 252) \\ an = 8 - 252 \\ an = - 244 \\ \\ \\ \geqslant soma \: dos \: termos \: da \: pa \\ \\ sn = \frac{(a1 + an)n}{2} \\ \\ sn = \frac{(8 + ( - 244)43}{2} \\ \\ sn = \frac{ - 236 \times 43}{2} \\ \\ sn = - 118 \times 43 \\ \\ sn = - 5074 \\ \\ \\ > < > < > < > < >$