Question

ache a soma dos 40 primeiros termos da p.a (10,2 )​

Answer 1

$> \: resolucao \\ \\ \geqslant \: progressao \: \: aritmetica \\ \\ r = a2 - a1 \\ r = 2 - 10 \\ r = - 8 \\ \\ = = = = = = = = = = = = = = = = \\ \\ > \: o \: 40 \: termo \: da \: pa \\ \\ an = a1 + (n - 1)r \\ an = 10 + (40 - 1) - 8 \\ an = 10 + 39 \times ( - 8) \\ an = 10 + ( - 312) \\ an = 10 - 312 \\ an = - 302 \\ \\ = = = = = = = = = = = = = = = = = = \\ \\ > \: soma \: dos \: 40 \: termos \: da \: pa \\ \\ sn = \frac{(a1 + an)n}{2} \\ \\ sn = \frac{(10 + ( - 302) \: )40}{2} \\ \\ sn = \frac{ - 292 \times 40}{2} \\ \\ sn = - 292 \times 20 \\ \\ sn = - 5840 \\ \\ \\ \geqslant \leqslant \geqslant \leqslant \geqslant \leqslant \geqslant \leqslant \geqslant$