Three non-zero real numbers form a AP and the squares of these numbers taken in same order form a GP. If the possible common ratios are `(2pm sqrt(k))` where `k in N`, then the value of `[k)/(8)-(8)/(k))` is (where [] denites the greatest integer function).

Explanation

Let number of AP are `(a-d),a,(a+d)`.
According to the question, `(a-d)^(2),a^(2),(a+d)^(2)` are in GP.
`:.(a^(2))^(2)=(a-d)^(2)(a+d)^(2)`
`implies a^(4)=(a^(2)-d^(2))^(2)`
`implies a^(4)=a^(4)+d^(4)-2a^(2)d^(2)`
`implies a^(2)(a^(2)-2d^(2))=0`
`implies ane0," So "a^(2)=2d^(2)`
`implies a=pm sqrt(2d)" " "......(i)"`
Let common ratio of GP is r.
`:.r^(2)=((a+d)^(2))/((a-d)^(2))`
`implies r^(2)=(a^(2)+d^(2)+2ad)/(a^(2)+d^(2)-2ad)`
`implies r^(2)=(2d^(2)+d^(2)+2sqrt(d^(2)))/(2d^(2)+d^(2)-2sqrt(d^(2)))`
`" " [" from Eq. (i) for "a= sqrt(2)d]`
`implies r^(2)=((3+2sqrt(2))d^(2))/((3-2sqrt(2))d^(2))`
`implies r^(2)=((3+2sqrt(2))(3+2sqrt(2)))/(9-8)`
`implies r^(2)=(3+2sqrt(2))^(2)`
`implies r^(2)=(3+sqrt(8))^(2)`
`:. r=pm(3+sqrt(8))`
`impliesr=3+sqrt(8) " " [:." r is positive "]`
Similarly, for `a=-sqrt(2)d,` we get
`r=pm(3-sqrt(8))`
`implies r=(3-sqrt(8)) " " [:. " r is positive "]`
Compare r with `3pmsqrt(k)`, we get
`k=8`
`[(k)/(8)-(8)/(k)]=[(8)/(8)-(8)/(8)]`
`= [1-1]=[0]=0`.