Question

a soma dos termos da P.A. (5+x,10+x,15+x,...100+x) e igual a 1110.qual e valor de x?

Answer 1

a1 = ( 5 + x)
a2 = ( 10 + x)
r = ( 10 + x) - ( 5 + x) = 10 + x - 5 - x = 5 *****
an = a1 + (n-1)r
( 100 + x) = ( 5 + x) + (n-1)5
100 + X = ( 5 + X ) + 5n - 5
100 + x = x + 5n
100 + x - x = 5n
5n = 100
n = 20 *****

S20 = ( a1 + a20 )* n/2
1110 = ( 5 + x + 100 + x ) * 10
( 2x + 105)* 10 = 1110
20x + 1050 = 1110
20x = 1110 - 1050
20x = 60
x = 60/20 = 3 ******

Answer 2

Os termos sobem de 5 em 5, exemplo: 15+x, 20+x, 25+x. Sendo assim 100/5= 20 termos.

$S_{20}= \frac{(A_{1}+A_{20})*20 }{2} -\ \textgreater \ S_{20}= (5+x+100+x)*10$
$1110= (105+2x)*10 =\ \textgreater \ 1110=1050+20x$
1110-1050=20x
x=$\frac{60}{20}$
x=3