If for the series a_1, a_2, a_3, ldots, the e | vedclass

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(C) Given that $\frac{a_r - a_{r+1}}{a_r a_{r+1}} = K$ (where $K$ is a constant).Dividing the terms, we get $\frac{a_r}{a_r a_{r+1}} - \frac{a_{r+1}}{

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$H$.$P$.

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(C) Given that $\frac{a_r - a_{r+1}}{a_r a_{r+1}} = K$ (where $K$ is a constant).Dividing the terms, we get $\frac{a_r}{a_r a_{r+1}} - \frac{a_{r+1}}{a_r a_{r+1}} = K$.This simplifies to $\frac{1}{a_{r+1}} - \frac{1}{a_r} = K$.Since the difference between the reciprocals of consecutive terms is constant, the sequence $\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \ldots$ is an $A$.$P$.Therefore, the sequence $a_1, a_2, a_3, \ldots$ is in $H$.$P$.

If for the series $a_1, a_2, a_3, \ldots$, the expression $a_r - a_{r+1}$ bears a constant ratio with $a_r a_{r+1}$, then $a_1, a_2, a_3, \ldots$ are in:

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