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Cartesian Product

If a1, a2...

Question

If $a_{1},$ $a_{2},$ $. . . . . a_{n}$ are in $H . P$ . then $a_{1} . a_{2} + a_{2} . a_{3} + a_{3} . a_{4} + . . . . . + a_{n - 1} . a_{n} =$

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Solution

Given, $a_{1}, a_{2}, a_{3}, . . . . . . . a_{n} - H . P$

$\frac{1}{a_{1}}, \frac{1}{a_{2}}, . . . . . . \frac{1}{a_{n}} - A . P$

$\frac{1}{a_{2}} - \frac{1}{a_{1}} = d, \frac{1}{a_{3}} - \frac{1}{a_{2}} = d, \frac{1}{a_{n}} - \frac{1}{a_{1}} = (n - 1) d - - - - - (A)$

$\frac{a_{1} - a_{2}}{a_{1} a 2} = \frac{a_{2} - a_{3}}{a_{2} a_{3}} = \frac{a_{3} - a_{4}}{a_{3} a_{4}} = . . . . . . . . = \frac{a_{n - 1} - a_{n}}{a_{n - 1} a_{n}} = d$

$a_{1} a_{2} = \frac{a_{1} - a_{2}}{d} . . . . . . . . . a_{n - 1} a_{n} = \frac{a_{n - 1} - a_{n}}{d}$

$= a_{1} a_{2} + a_{2} + a_{3} + . . . . . . . . a_{n - 1} a_{n}$

$= \frac{a_{1} - a_{2}}{d} + \frac{a_{2} - a_{3}}{d} + . . . . . . \frac{a_{n - 1} - a_{n}}{d}$

$= \frac{a_{1} - a_{n}}{d}$

$= (n - 1) a \cdot a_{n}$ .................. From $(A)$

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a_{1}, a_{2}, a_{3}, . . . . . . a_{n}

are in H.P then

a_{1} a_{2} + a_{2} a_{3} + a_{3} a_{4} . . . . + a_{n - 1} a_{n} =

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are in H.P then

\frac{a_{1}}{a_{2} + a_{3} + . . . a_{n}}, \frac{a_{2}}{a_{1} + a_{3} + . . . a_{n}}, \frac{a_{3}}{a_{1} + a_{2} + a_{4} + . . . + a_{n}}, \frac{a_{n}}{a_{1} + a_{2} + . . . + a_{n}}

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