Question

# If a1,a2,a3…,an are in AP, where ai>0 for all i, then the value of 1√a1+√a2+1√a2+√a3+…+1√an−2+√an is

A
1a1+an
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B
1a1an
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C
na1+an
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D
(n1)a1+an
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Solution

## The correct option is D (n−1)√a1+√an1√a1+√a2+1√a2+√a3+…+1√an−1+√an=1√a1+√a2×√a1−√a2√a1−√a2+1√a2+√a3×√a2−√a3√a2−√a3+…+1√a1+√a2×√an−1−√an√an−1−√an=√a1−√a2√a1−√a2+√a2−√a3√a2−√a3+…+√an−1−√an√an−1−√an=√a1−√a2−d+=√a2−√a3−d+…+√an−1−√an−d [d= common difference] =√a2−√a1d+√a3−√a2d+…+√an−√an−1d=1d[√a2−√a1+√a3−√a2+…+√an−√an−1]=1d[−√a1+√an] (Rest all terms will cancelled) =1d[−√a1+√an]=1d[√an−√a1]=1d√an−√a1×√an+√a1√an+√a1=1dan−a1√an+√a1=1da1+(n−1)d−a1√an+√a1=1d(n−1)d√an+√a1=n−1√a1+√an

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