Q.

If α,β,γ,and δare in arithmetic progression.  Then sinα+β+γ+δ≠

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a

sin3β+δ

b

sinα+3γ

c

sin2α+2δ

d

sin3β+γ

answer is D.

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Detailed Solution

Let   β=α+d,  γ=α+2d,  δ=α+3d          ⇒ α+β+γ+δ = 4α+6d →1Now          3β+δ=3α+d+α+3d=4α+6d→2By 1  and 2 , we have   α+β+γ+δ=          3β+δ       Also α+3γ=α+3α+2d = 4α+6d→3       By 1 and 3 we have α+β+γ+δ=α+3γα+β+γ+δ=2α+2δ
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