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,$\frac{1}{a_{k+1}} - \frac{1}{a_k} = d$ for $k = 1, 2, \dots, n-1$.This implies $\frac{a_k - a_{k+1}}{a_k a_{k+1}} = d$,or $a_k a_{k+1} = \frac{1}{d}(a_k - a_{k+1})$.Summing this from $k=1$ to $n-1$:$\sum_{k=1}^{n-1} a_k a_{k+1} = \frac{1}{d} \sum_{k=1}^{n-1} (a_k - a_{k+1}) = \frac{1}{d} (a_1 - a_n)$.For the $A.P.$,the $n^{th}$ term is $\frac{1}{a_n} = \frac{1}{a_1} + (n-1)d$,so $d = \frac{a_1 - a_n}{(n-1)a_1 a_n}$.Substituting $d$ into the sum:$\sum_{k=1}^{n-1} a_k a_{k+1} = \frac{1}{\frac{a_1 - a_n}{(n-1)a_1 a_n}} (a_1 - a_n) = (n-1)a_1 a_n$.

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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