Question

Interpole 9 meios aritméticos entre os números 10 e 98.​

Answer 1

Interpolando 9 meios aritméticos entre os dois já existentes, a PA ficará com um total de 11 termos.

10 , a2 , a3 , a4 , a5 , a6 , a7 , a8 , a9 , a10 , 98

Vamos determinar a razão utilizando a equação do termo geral da PA:

$a_n~=~a_1+(n-1).r\\\\\\a_{11}~=~a_1+(11-1).r\\\\\\98~=~10+10r\\\\\\10r~=~98-10\\\\\\r~=~\frac{88}{10}\\\\\\\boxed{r~=~8,8}$

Podemos agora determinar os termos inseridos na PA:

$a_2~~=~~a_1+r~~=~~10+8,8~~~~\rightarrow~~\boxed{a_2~=~18,8}\\\\a_3~~=~~a_2+r~~=~~18,8+8,8~~~~\rightarrow~~\boxed{a_3~=~27,6}\\\\a_4~~=~~a_3+r~~=~~27,6+8,8~~~~\rightarrow~~\boxed{a_4~=~36,4}\\\\a_5~~=~~a_4+r~~=~~36,4+8,8~~~~\rightarrow~~\boxed{a_5~=~45,2}\\\\a_6~~=~~a_5+r~~=~~45,2+8,8~~~~\rightarrow~~\boxed{a_6~=~54}\\\\a_7~~=~~a_6+r~~=~~54+8,8~~~~\rightarrow~~\boxed{a_7~=~62,8}$

$a_8~~=~~a_7+r~~=~~62,8+8,8~~~~\rightarrow~~\boxed{a_8~=~71,6}\\\\a_9~~=~~a_8+r~~=~~71,6+8,8~~~~\rightarrow~~\boxed{a_9~=~80,4}\\\\a_{10}~~=~~a_9+r~~=~~80,4+8,8~~~~\rightarrow~~\boxed{a_{10}~=~89,2}$