Question

A soma dos 20 primeiros termos da P.A (-10,-6,-2…)é?

Answer 1

$> \: resolucao \\ \\ \geqslant \: progressao \: \: aritmetica \\ \\ r = a2 - a1 \\ r = - 6 - ( - 10) \\ r = - 6 + 10 \\ r = 4 \\ \\ = = = = = = = = = = = = = = = = = \\ \\ > \: o \: 20 \: termo \: da \: pa \\ \\ an = a1 + (n - 1)r \\ an = - 10 + (20 - 1)4 \\ an = - 10 + 19 \times 4 \\ an = - 10 + 76 \\ an = 66 \\ \\ = = = = = = = = = = = = = = = = = \\ \\ > \: a \: soma \: dos \: termos \: da \: pa \\ \\ sn = \frac{(a1 + an)n}{2} \\ \\ sn = \frac{( - 10 + 66)20}{2} \\ \\ sn = \frac{56 \times 20}{2} \\ \\ sn = 56 \times 10 \\ \\ sn = 560 \\ \\ \\ \geqslant \leqslant \geqslant \leqslant \geqslant \leqslant \geqslant \geqslant$

Answer 2

Resposta:

$\textsf{Leia abaixo}$

Explicação passo-a-passo:

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

$\mathsf{a_{20}=-10+(20-1)\cdot4 }$

$\mathsf{ a_{20}=-10+19\cdot4}$

$\mathsf{a_{20}=-10+76 }$

$\mathsf{a_{20}=66 }$

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

$\mathsf{S_{20}=\dfrac{(-10+66)\cdot20}{2} }$

$\mathsf{S_{20}=\dfrac{56\cdot20}{2} }$

$\mathsf{S_{20}=28\cdot20 }$

$\boxed{\boxed{ \mathsf{S_{20}=560}} }$