Three integers are in the ratio 2:3:8. If 4 is added to the middle number, the resulting number is the second term of a geometric progression
Can anyone please solve this problem?
thanks in advance
isn't that middle term a mean?
im try:
2x, 3x, 8x
let x be the unknown number
consider m be the mean then 2x/m = m/8x
m = sq. rt of 16x^2
m = 4x this is the mean or middle.
so, 4x + 4 = 3x
x = - 4 ...
this substitute...
hope this is ryt . correct me if i'm wrong.. thanks