Given that - - a - b are in A-P- - - c - d are in G-P- and - - e - f are in H-P- If b - d - f are in G-P- then -6-4-4 equals toA- 3-2B-C-D-

The correct option is A 3/2α, β, a, b are in A.P b=a+3(β α )=3β 2α α,β, c, d are in G.P. d=α ( β/α )^3=β ^3/α ^2 α , β, e, f are in H.P 1/f=1/α+3 ( 1/β 1/α ) ...

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Byju's Answer

Standard XII

Mathematics

Chain Rule of Differentiation

Given that α,...

Question

Given that $α, β, a, b$ are in A.P., $α, β, c, d$ are in G.P. and $α, β, e, f$ are in H.P. If b, d, f are in G.P., then $\frac{β^{-} α^{6}}{α β (β^{4} - α^{4})}$ equals to

\frac{2}{3}

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\frac{3}{2}

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\frac{4}{3}

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\frac{3}{4}

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Solution

The correct option is B $\frac{3}{2}$
$α, β, a, b$ are in A.P
$b = a + 3 (β - α) = 3 β - 2 α$
$α, β, c, d$ are in G.P.
$d = α {(\frac{β}{α})}^{3} = \frac{β^{3}}{α^{2}}$
$α, β, e, f$ are in H.P
$\frac{1}{f} = \frac{1}{α} + 3 (\frac{1}{β} - \frac{1}{α})$
$f = \frac{α β}{3 α - 2 β}$
Now since b, d, f are in G.P.
$d^{2} = b f$
${(\frac{β^{3}}{α^{2}})}^{2} = (3 β - 2 α) \frac{(α β)}{3 α - 2 β}$
$β^{5} (3 α - 2 β) = α^{5} (3 β - 2 α)$
$3 α β (β^{4} - α^{4}) = 2 (β^{6} - α^{6})$
$\frac{3}{2} = \frac{β^{6} - α^{6}}{α β (β^{4} - α^{4})}$

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Given that α,β,a,b are in A.P., α,β,c,d are in G.P. and α,β,e,f are in H.P. If b, d, f are in G.P., then β−α6αβ(β4−α4) equals to

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