If {a_1}, {a_2}, dots, {a_{n+1}} are in A.P., | vedclass

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(D) Given that ${a_1}, {a_2}, \dots, {a_{n+1}}$ are in $A.P.$ with common difference $d = {a_{k+1}} - {a_k}$.Let $S = \sum_{k=1}^{n} \frac{1}{{{a_k}{a

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$\frac{n}{{{a_1}{a_{n+1}}}}$

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(D) Given that ${a_1}, {a_2}, \dots, {a_{n+1}}$ are in $A.P.$ with common difference $d = {a_{k+1}} - {a_k}$.Let $S = \sum_{k=1}^{n} \frac{1}{{{a_k}{a_{k+1}}}}$.Since ${a_{k+1}} - {a_k} = d$,we can write $\frac{1}{{{a_k}{a_{k+1}}}} = \frac{1}{d} \left( \frac{{a_{k+1}} - {a_k}}{{{a_k}{a_{k+1}}}} \right) = \frac{1}{d} \left( \frac{1}{{{a_k}}} - \frac{1}{{{a_{k+1}}}} \right)$.Substituting this into the sum:$S = \frac{1}{d} \left[ \left( \frac{1}{{{a_1}}} - \frac{1}{{{a_2}}} \right) + \left( \frac{1}{{{a_2}}} - \frac{1}{{{a_3}}} \right) + \dots + \left( \frac{1}{{{a_n}}} - \frac{1}{{{a_{n+1}}}} \right) \right]$.This is a telescoping series,so all intermediate terms cancel out:$S = \frac{1}{d} \left( \frac{1}{{{a_1}}} - \frac{1}{{{a_{n+1}}}} \right) = \frac{1}{d} \left( \frac{{{a_{n+1}} - {a_1}}}{{{a_1}{a_{n+1}}}} \right)$.Since ${a_{n+1}} = {a_1} + nd$,we have ${a_{n+1}} - {a_1} = nd$.Therefore,$S = \frac{1}{d} \left( \frac{nd}{{{a_1}{a_{n+1}}}} \right) = \frac{n}{{{a_1}{a_{n+1}}}}$.Thus,the correct option is $D$.

If ${a_1}, {a_2}, \dots, {a_{n+1}}$ are in $A.P.$,then $\frac{1}{{{a_1}{a_2}}} + \frac{1}{{{a_2}{a_3}}} + \dots + \frac{1}{{{a_n}{a_{n+1}}}}$ is

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