If a 1 - a 2 - a 3 - a 4 are in AP- then 1 - a 1 - a 2 - 1 - a 2 - a 3 - 1 - a 3 - a 4 is equal to

The correct option is A √( a _ 4 ) √( a _ 1 ) a _ 3 a _ 2 1 √( a _ 1 ) +√( a _ 2 ) #160; + 1 √( a _ 2 ) +√( a _ 3 ) #160; + 1 √( a _ 3 ) +√( a ...

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Sum of n Terms

If a 1 , a ...

Question

If $a_{1}, a_{2}, a_{3}, a_{4}$ are in AP, then
$\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \frac{1}{\sqrt{a_{3}} + \sqrt{a_{4}}}$ is equal to

\frac{\sqrt{a_{4}} - \sqrt{a_{1}}}{a_{3} - a_{2}}

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\frac{a_{4} - a_{1}}{a_{3} - a_{2}}

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\frac{a_{3} - a_{2}}{\sqrt{a_{4}} - \sqrt{a_{1}}}

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\frac{a_{1} - a_{4}}{a_{3} - a_{1}}

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\frac{a_{5} - a_{0}}{a_{1} - a_{4}}

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a_{1}, a_{2}, a_{3}, a_{4}, a_{5} > 0

\frac{a_{1}}{a_{2} + a_{3}} + \frac{a_{2}}{a_{3} + a_{4}} + \frac{a_{3}}{a_{4} + a_{5}} + \frac{a_{4}}{a_{5} + a_{1}} + \frac{a_{5}}{a_{1} + a_{2}} \geq \frac{5}{2}

a_{1}, a_{2}, a_{3}, a_{4}, . . {. ., a}_{20}

a_{1} + a_{20} = 45

a_{1} + a_{2} + a_{3} + a_{4}, . . {. . + a}_{20}

a_{1}, a_{2}, a_{3}, a_{4}

a_{1}

a_{2}

a_{3}

a_{2}

a_{3}

a_{4}

a_{3}

a_{4}

a_{5}

a_{1}

a_{3}

a_{5}

\frac{a_{2} a_{3}}{a_{1} a_{4}} = \frac{a_{2} + a_{3}}{a_{1} + a_{4}} = 3 (\frac{a_{2} - a_{3}}{a_{1} - a_{4}})

a_{1}

a_{2}

a_{3}

a_{4}

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