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

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- March 2nd 2011, 08:22 PMjam2011geometric progression
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 - March 2nd 2011, 10:02 PMrcs
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 - March 2nd 2011, 10:34 PMsa-ri-ga-ma
**4x + 4 = 3x**

It should be 4x = 3x + 4.