Question

If the first and the \((2 n+1)^{\text {th }}\) t erms of an AP, GP and HP are equal and their \(n^{\text {th }}\) terms are respectively a, b, c then always
(A) \(\mathrm{a}=\mathrm{b}=\mathrm{c}\)
(B) \(a \geq b \geq c\)
(C) \(\mathrm{a}+\mathrm{c}=\mathrm{b}\)
(D) \(a c-b^{2}=0\)

Answer 1

Ans : \((\mathbf{B}, \mathbf{D})\)
 There seems to be a printing mistake here
If there are \((2 n-1)\) terms instead of \((2 n+1)\) terms then \(n^{\text {th }}\) terms of the A.P., G.P. and H.P. are the A.M., G.M. \& H.M
of the first and the last terms.
So, \(a \geq b \geq c \& a c-b^{2}(B, D)\)
otherwise if there are \((2 n+1)\) terms then the \(n^{\text {th }}\) terms should be in decreasing order of A.P., G.P. \& H.P.
i.e. \(\mathbf{a} \geq \mathbf{b} \geq \mathbf{c}\). (B)

Answer 2

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Answer 3

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