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 the value of -6-6-4-4 is

Α,β, a,b are in A.P. ⇒ b=α+3(β α)=3β 2α α,β,c,d are in G.P. ⇒ d=α·(βα)^3=β^3α^2 α,β,e,f are in H.P. ⇒1f=1α+3(1β 1α) ⇒ f=αβ3α 2β Now, since b,d,f are in G.P., ...

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Byju's Answer

Standard XII

Mathematics

Sum of Infinite Terms of a GP

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 the value of $\frac{β^{6} - α^{6}}{α β (β^{4} - α^{4})}$ is

\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.
$\Rightarrow b = α + 3 (β - α) = 3 β - 2 α$

$α, β, c, d$ are in G.P.
$\Rightarrow d = α \cdot {(\frac{β}{α})}^{3} = \frac{β^{3}}{α^{2}}$

$α, β, e, f$ are in H.P.
$\Rightarrow \frac{1}{f} = \frac{1}{α} + 3 (\frac{1}{β} - \frac{1}{α})$
$\Rightarrow f = \frac{α β}{3 α - 2 β}$

Now, since $b, d, f$ are in G.P.,
$d^{2} = b f$
$\Rightarrow {(\frac{β^{3}}{α^{2}})}^{2} = (3 β - 2 α) \frac{α β}{(3 α - 2 β)}$
$\Rightarrow β^{5} (3 α - 2 β) = α^{5} (3 β - 2 α)$
$\Rightarrow 3 α β^{5} - 2 β^{6} = 3 β α^{5} - 2 α^{6}$
$\Rightarrow 3 α β (β^{4} - α^{4}) = 2 (β^{6} - α^{6})$
$\Rightarrow \frac{β^{6} - α^{6}}{α β (β^{4} - α^{4})} = \frac{3}{2}$

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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 the value of β6−α6αβ(β4−α4) is

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