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 2-6-6-4-4- 0- is

Given that α, β, a, b are in A.P. ⇒ b α=3(β α) ⇒ b=3β 2α....(1) Also, nbsp;α, β, c, d are in G.P. ⇒dα=(βα)^3 ⇒ d=β^3α^2.....(2) Similarly, using the l ...

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Byju's Answer

Standard XIII

Mathematics

nth Term of HP

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 $\frac{2 (α^{6} - β^{6})}{α β (α^{4} - β^{4})}, 0 < α < β$ is

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Solution

Given that $α, β, a, b$ are in A.P.
$\Rightarrow b - α = 3 (β - α) \Rightarrow b = 3 β - 2 α . . . . (1)$

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

Similarly, using the last condition, we get
$\frac{1}{f} = \frac{3}{β} - \frac{2}{α} . . . . . . (3)$

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

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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 2(α6−β6)αβ(α4−β4),0<α<β is

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