Let (α,β,γ) and Q(1,−1,0) be two points such that the mid-point of PQ is R(x,y,z). If x is the A.M. of α and β, y is the G.M. of β and γ rand z is the H.M. of γ and α, then
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Solution
Since R is the midpoint of PQ, we have 2x=α+1,2y=β−1,2z=γ Also, 2x=α+β since x is the A.M. of α & β. y2=β×γ z=2α×γα+γ Using the above relations, β=1. Also, 4α=α+γ, or γ=3α Also, since β is 1,y=0. Hence, γ becomes 0, implying α to be 0. So, α+β+γ=1