Question

determine a soma dos termos da p.a. (4,7,10,...,136) .​

Answer 1

$> resolucao \\ \\ \geqslant progressao \: aritmetica \\ \\ r = a2 - a1 \\ r = 7 - 4 \\ r = 3 \\ \\ \\ an = a1 + (n - 1)r \\ 136 = 4 + (n - 1)3 \\ 136 = 4 + 3n - 3 \\ 136 = 1 + 3n \\ 136 - 1 = 3n \\ 135 = 3n \\ n = \frac{135}{3} \\ n = 45 \\ \\ \geqslant \: soma \: dos \: termos \: da \: pa \\ \\ \\ sn = \frac{(a1 + an)n}{2} \\ \\ sn = \frac{(4 + 136)45}{2} \\ \\ sn = \frac{140 \times 45}{2} \\ \\ sn = 70 \times 45 \\ \\ sn = 3150 \\ \\ \\ > < > < > < > < > < > < > >$