If {a_1}, {a_2}, {a_3}, dots, {a_n} are in H. | vedclass

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(C) Since ${a_1}, {a_2}, {a_3}, \dots, {a_n}$ are in $H.P.$,their reciprocals $\frac{1}{a_1}, \frac{1}{a_2}, \dots, \frac{1}{a_n}$ are in $A.P.$Let th

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$(n - 1){a_1}{a_n}$

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(C) Since ${a_1}, {a_2}, {a_3}, \dots, {a_n}$ 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 of this $A.P.$ be $d$.Then,$\frac{1}{a_{k+1}} - \frac{1}{a_k} = d$,which implies $\frac{a_k - a_{k+1}}{a_k a_{k+1}} = d$,or $a_k - a_{k+1} = d(a_k a_{k+1})$.Summing this from $k=1$ to $n-1$:$\sum_{k=1}^{n-1} (a_k - a_{k+1}) = d \sum_{k=1}^{n-1} (a_k a_{k+1})$$(a_1 - a_2) + (a_2 - a_3) + \dots + (a_{n-1} - a_n) = d \sum_{k=1}^{n-1} (a_k a_{k+1})$$a_1 - a_n = d \sum_{k=1}^{n-1} (a_k a_{k+1})$For the $A.P.$,the $n^{th}$ term is $\frac{1}{a_n} = \frac{1}{a_1} + (n-1)d$,so $d = \frac{1/a_n - 1/a_1}{n-1} = \frac{a_1 - a_n}{(n-1)a_1 a_n}$.Substituting $d$ into the sum equation:$a_1 - a_n = \frac{a_1 - a_n}{(n-1)a_1 a_n} \sum_{k=1}^{n-1} (a_k a_{k+1})$$\sum_{k=1}^{n-1} (a_k a_{k+1}) = (n-1)a_1 a_n$.

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

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