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प्रश्न
If a, b, c is three consecutive terms of an A.P. and x, y, z are three consecutive terms of a G.P. then prove that xb–c × yc–a × za–b = 1
उत्तर
a, b, c is three consecutive terms of an AP.
∴ Let a, b, c be a, a + d, a + 2d respectively ...(1)
x, y, z is three consecutive terms of a GP.
∴ Assume x, y, z as x, x.r, x.r2 respectively ...(2)
PT : xb–c, yc–a, za–b = 1
Substituting (1) and (2) in L.H.S., we get
L.H.S. = `x^("a" + "d" - "a" - 2"d") xx (x"r")^("a" + 2"d" - "a") xx (x"r"^2)^("a" - "a" - "d")`
= (x)–d . (xr)2d (xr2)–d
= `1/x^4 xx x^(2"d") * "r"^(2"d") xx 1/(x^"d" "r"^(2"d")` = 1 = R.H.S.
Hence proved.
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