Question

18) obter a razão da pa em que o primeiro termo e -8 eo vigésimo e 30 19) obter a soma dos 12 primeiros termos de uma pa (6,14,22,…) 20) calcule a soma dos 10 primeiros termos da pa (38,42,46

Answer 1

Vamos dividir o número da do termo com o valor do termo

$\mathtt{R = \dfrac{A_{20} - A_1}{A_{20}- A_1}~~=~~ \dfrac{30-(-8)}{20-1}~~=~~\dfrac{30 +8}{19}~~=~~\dfrac{38}{19}~~=2}$

19) (6,14,22,…)

$\mathtt{R = A_2 - A_1} \\ \mathtt{R = 14 - 6} \\ \mathtt{R =8 } \\ \\ \mathtt{A_n = A_1 + (n-1)~.~R} \\ \mathtt{A_{12} = 6 + (12-1)~.~8} \\ \mathtt{A_{12} = 6 + 11~.~8} \\ \mathtt{A_{12} = 6 + 88} \\ \mathtt{A_{12} = 94}} \\ \\ \\ \mathtt{S_{n} = \dfrac{(A_n + A_1)~.~n}{2}} \\ \\ \\ \mathtt{S_{12} = \dfrac{(94 + 6 )~.~12}{2}~~=~~\dfrac{100~.~12}{2}~~=~~\dfrac{1.200}{2}~~=~~600}$

20) (38,42,46..)

$\mathtt{R = A_2 - A_1} \\ \mathtt{R = 42 - 38} \\ \mathtt{R =4 } \\ \\ \\ \mathtt{A_n = A_1 + (n-1)~.~R} \\ \mathtt{A_{10} = 38 + (10-1)~.~4} \\ \mathtt{A_{10} = 38 + 9 ~.~4} \\ \mathtt{A_{10} = 38 +36} \\ \mathtt{A_{10} = 74} \\ \\ \\ \mathtt{S_n = \dfrac{(A_n + A_1)~.~n}{2}} \\ \\ \\ \mathtt{S_{10} = \dfrac{(74 + 38)~.~10}{2}~~=~~\dfrac{112~.~10}{2}~~=~~\dfrac{1.120}{2}~~=~~560}$