If a_1, a_2, a_3, dots, a_n are in H.P., then | vedclass

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(C) Since $a_1, a_2, \dots, a_n$ are in $H.P.$, their reciprocals $\frac{1}{a_1}, \frac{1}{a_2}, \dots, \frac{1}{a_n}$ are in $A.P.$Let the common dif

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$(n - 1) a_1 a_n$

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(C) Since $a_1, a_2, \dots, a_n$ are in $H.P.$, their reciprocals $\frac{1}{a_1}, \frac{1}{a_2}, \dots, \frac{1}{a_n}$ are in $A.P.$Let the common difference be $d = \frac{1}{a_{k+1}} - \frac{1}{a_k}$.Then, $a_k a_{k+1} = \frac{1}{d} (a_k - a_{k+1})$.The given sum is $S = \sum_{k=1}^{n-1} a_k a_{k+1} = \sum_{k=1}^{n-1} \frac{a_k - a_{k+1}}{d} = \frac{1}{d} (a_1 - a_n)$.From the $A.P.$ property, $\frac{1}{a_n} = \frac{1}{a_1} + (n-1)d$, which implies $\frac{1}{a_n} - \frac{1}{a_1} = (n-1)d$.Thus, $\frac{a_1 - a_n}{a_1 a_n} = (n-1)d$, so $\frac{a_1 - a_n}{d} = (n-1) a_1 a_n$.Therefore, the sum is $(n-1) a_1 a_n$.

If $a_1, a_2, a_3, \dots, a_n$ are in $H.P.$, then the expression $a_1 a_2 + a_2 a_3 + \dots + a_{n-1} a_n$ is equal to:

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