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$If x_{1}, x_{2}, \dots, x_{n} are in H.P, then \sum_{r = 1}^{n - 1} x_{r} x_{r + 1} is equal to$

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a

n−1x1xn

b

nx1xn

c

n+1x1xn

d

nx1x2

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detailed solution

Correct option is A

Clearly 1x1,1x2,−−−1xn will be in A.P ⇒1x2−1x1=1x3−1x2=−−−=1xr+1−1xr=−−⇒1xn−1xn−1=λ(say)⇒∑r=1n−1 xrxr+1=1λ∑r=1n−1 xr−xr+1=1λx1−xn Now __,1xn=1x1+(n−1)λ⇒x1−xnx1xn=(n−1)λ=∑r=1n−1 xrxr+1=(n−1)x1xn

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Given that $α, γ$ are roots of the equation $A x^{2} - 4 x + 1 = 0,$ and $β, δ$ the roots of the equation of $B x^{2} - 6 x + 1 = 0,$ such that $α, β, γ,$ and $δ$ are in H.P., then

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