First slide

Harmonic Progression

Question

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

Difficult

A

$(n - 1) x_{1} x_{n}$
B

$n x_{1} x_{n}$
C

$(n + 1) x_{1} x_{n}$
D

$n x_{1} x_{2}$

Solution

$\begin{array}{l} Clearly \frac{1}{x_{1}}, \frac{1}{x_{2}}, - - - \frac{1}{x_{n}} will be in A.P \\ \Rightarrow \frac{1}{x_{2}} - \frac{1}{x_{1}} = \frac{1}{x_{3}} - \frac{1}{x_{2}} = - - - = \frac{1}{x_{r + 1}} - \frac{1}{x_{r}} = - - \\ \Rightarrow \frac{1}{x_{n}} - \frac{1}{x_{n - 1}} = λ (s a y) \\ \Rightarrow \sum_{r = 1}^{n - 1} x_{r} x_{r + 1} = \frac{1}{λ} \sum_{r = 1}^{n - 1} (x_{r} - x_{r + 1}) = \frac{1}{λ} (x_{1} - x_{n}) \\ \underline{\underline{Now}}, \frac{1}{x_{n}} = \frac{1}{x_{1}} + (n - 1) λ \Rightarrow \frac{x_{1} - x_{n}}{x_{1} x_{n}} = (n - 1) λ \\ = \sum_{r = 1}^{n - 1} x_{r} x_{r + 1} = (n - 1) x_{1} x_{n} \end{array}$

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