Math miscellaneousMedium
Question
If x, y, z are in A.P., then 3kx+1, 3ky+1, 3kz+1 are in (k ≠ 0)
Options
A.A.P
B.G.P.
C.H.P.
D.None of these
Solution
3kx+1, 3ky+1, 3kz+1
⇒ 32(ky + 1) = 3k(x + z) + 2 ⇒ 2ky + 2 = kx + kz + 2
⇒ 2y = x + z ⇒ 3kx + 1, 3ky + 1, 3kz + 1 are in G.P
⇒ 32(ky + 1) = 3k(x + z) + 2 ⇒ 2ky + 2 = kx + kz + 2
⇒ 2y = x + z ⇒ 3kx + 1, 3ky + 1, 3kz + 1 are in G.P
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