If \[{{a}_{1}},\,{{a}_{2}},....,{{a}_{n+1}}\] are in A.P., then \[\frac{1}{{{a}_{1}}{{a}_{2}}}+\frac{1}{{{a}_{2}}{{a}_{3}}}+.....+\frac{1}{{{a}_{n}}{{a}_{n+1}}}\] is [AMU 2002]

\[\frac{n-1}{{{a}_{1}}{{a}_{n+1}}}\]

\[\frac{1}{{{a}_{1}}{{a}_{n+1}}}\]

\[\frac{n+1}{{{a}_{1}}{{a}_{n+1}}}\]

\[\frac{n}{{{a}_{1}}{{a}_{n+1}}}\]

If \[{{a}_{1}},\,{{a}_{2}},....,{{a}_{n+1}}\] are in A.P., then \[\fra

1.	If \[{{a}_{1}},\,{{a}_{2}},....,{{a}_{n+1}}\] are in A.P., then \[\frac{1}{{{a}_{1}}{{a}_{2}}}+\frac{1}{{{a}_{2}}{{a}_{3}}}+.....+\frac{1}{{{a}_{n}}{{a}_{n+1}}}\] is [AMU 2002]
A.	\[\frac{n-1}{{{a}_{1}}{{a}_{n+1}}}\]
B.	\[\frac{1}{{{a}_{1}}{{a}_{n+1}}}\]
C.	\[\frac{n+1}{{{a}_{1}}{{a}_{n+1}}}\]
D.	\[\frac{n}{{{a}_{1}}{{a}_{n+1}}}\]
Answer» E.

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If \[{{a}_{1}},\,{{a}_{2}},....,{{a}_{n+1}}\] are in A.P., then \[\frac{1}{{{a}_{1}}{{a}_{2}}}+\frac{1}{{{a}_{2}}{{a}_{3}}}+.....+\frac{1}{{{a}_{n}}{{a}_{n+1}}}\] is [AMU 2002]

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