If frac{a_2 a_3}{a_1 a_4} = frac{a_2 + a_3}{a | vedclass

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(C) Given $\frac{a_2 a_3}{a_1 a_4} = \frac{a_2 + a_3}{a_1 + a_4}$,we can write $\frac{a_1 + a_4}{a_1 a_4} = \frac{a_2 + a_3}{a_2 a_3}$.This implies $\

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Harmonic Progression

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(C) Given $\frac{a_2 a_3}{a_1 a_4} = \frac{a_2 + a_3}{a_1 + a_4}$,we can write $\frac{a_1 + a_4}{a_1 a_4} = \frac{a_2 + a_3}{a_2 a_3}$.This implies $\frac{1}{a_4} + \frac{1}{a_1} = \frac{1}{a_3} + \frac{1}{a_2}$,or $\frac{1}{a_4} - \frac{1}{a_3} = \frac{1}{a_2} - \frac{1}{a_1} \quad (1)$.Also,from $\frac{a_2 a_3}{a_1 a_4} = 3\left( \frac{a_2 - a_3}{a_1 - a_4} \right)$,we have $\frac{a_1 - a_4}{a_1 a_4} = 3\left( \frac{a_2 - a_3}{a_2 a_3} \right)$.This implies $\frac{1}{a_4} - \frac{1}{a_1} = 3\left( \frac{1}{a_3} - \frac{1}{a_2} \right) \quad (2)$.Let $x_n = \frac{1}{a_n}$. Then $(1)$ becomes $x_4 - x_3 = x_2 - x_1 = d$ (common difference).Substituting into $(2)$,$x_4 - x_1 = 3(x_3 - x_2)$.Since $x_4 - x_1 = (x_4 - x_3) + (x_3 - x_2) + (x_2 - x_1) = 3d$,and $x_3 - x_2 = d$,this is consistent.Thus,$\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \frac{1}{a_4}$ are in Arithmetic Progression.Therefore,$a_1, a_2, a_3, a_4$ are in Harmonic Progression.

If $\frac{a_2 a_3}{a_1 a_4} = \frac{a_2 + a_3}{a_1 + a_4} = 3\left( \frac{a_2 - a_3}{a_1 - a_4} \right)$,then in which progression are $a_1, a_2, a_3, a_4$?

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