The values of `xyz` is `15/2` or `18/5` according as the series `a, x, y, z, b` is an `AP` or `HP.` Find the values of `a & b` assuming them to be positive integer.

Explanation

`xyz=15/2` if a,x,y,z,b are in A.P. ltbr. Common difference = d=`(b-a)/(n+1)=(b-a)/4`
`rArrx=(3a+b)/4,y=(a+b)/2andz=(3b+a)/4`
`rArrxyz=15/2=((3a+b)/4)((a+b)/2)((3b+a)/4)` …(1)
Also xyz`=18/5` if a,x,y,z,b are in H.P.
`rArr1/a,1/x1/y,1/y,1/z,1/b` are in A.P.,
`rArrx=(4ab)/(3b+4),y=(2ab)/(a+b)andz=(4ab)/(3a+b)`
`rArrxyz=18/5=((4ab)/(3b+a))((2ab)/(a+b))((4ab)/(3a+b))` ...(2)
Multiplying equations (1) and (2), we get
`a^(3)b^(3)=18/5xx15/2=27`
`rArrab=3`
`rArra=1,3` or b=3,1