The value of xyz is 55 or `(343)/(55)` according as the series `a,x,y,z,b` is an AP or HP. Find the values of a and b given that they are positive integers.

Explanation

If `a,x,y,z,b` are in AP.
Then, b=Fifth term `=a+(5-1)d`
where dis common difference
`:. d=(b-a)/(4)`
`:.x*y*z=(a+d)(a+2d)(a+3d)=55 " " [" given "]`
`implies ((b+3a)/(4))((2a+2b)/(4))((a+3b)/(4))=55`
`implies (a+3b)(a+b)(3a+b)=55xx32 " " ".....(i)"`
If they are in HP.
The common difference oe the associated AP is `(1)/(4)((1)/(b-(1)/(a))`
`i.e., ((a-b))/(4ab)`
`:. (1 )/(x)=(1)/(a)+(2(a-b))/(4ab)`
`implies x=(4ab)/(a+3b)`
`:. (1)/(y)=(1)/(a)+(2(a-b))/(4ab)`
` implies y=(4ab)/(2a+2b)=(2ab)/(a+b)`
and `(1)/(z)=(1)/(a)+(3(a-b))/(4ab)`
`implies z=(4ab)/(3a+b)`
`:.xyz=(4ab)/((a+3b))*(2ab)/((a+b))*(4ab)/((3a+b))=343 " " [" given "]`
`implies (32a^(3)b^(3))/(55xx32)=(343)/(55) " " [" from Eq. (i) "]`
or `a^(3)b^(3)=343`
`implies ab=7`
Hence, `a=7,b=1`
or `a=1,b=7`.