If `a ,b ,c ,d` are in A.P. and `x , y , z` are in G.P., then show that `x^(b-c)doty^(c-a)dotz^(a-b)=1.`

Answer

N/a

Explanation

a, b, c are in A.P.
`rArr" "2b=a+c" ...(1)"`
x,y, z are in G.P.
`rArr" "y^(2)-xz" ...(2)"`
`L.H.S.=x^(b-c).y^(c-a).z^(a-b)`
`=x^(b-c).y^(c-a).((y^(2))/x)^(a-b)`
[From eq. (2)]
`=x^(b-c).y^(c-a).(y^(2a-2b))/(x^(a-b))`
`=x^(b-c-a+b).y^(c-a+2a-2b)`
`=x^(2b-(c+a)).y^((c+a)-2b)`
`=x^(2b-2b).y^(2b-2b)" [From eq. (1)]"`
`=x^(0).y^(0)=1=R.H.S`