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If a1,a2,...a...

Question

If $a_{1}, a_{2}, . . . a_{50}$ are in GP, then $(a_{1} - a_{3} + a_{5} - . . . + a_{49}) / (a_{2} - a_{4} + a_{6} - . . . + a_{50})$ is equal to

A

$a_{2} / a_{5}$

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B

$a_{4} / a_{6}$

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C

$a_{1} / a_{2}$

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D

$a_{6} / a_{1}$

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Solution

The correct option is C
$a_{1} / a_{2}$

Finding the value of $(a_{1} - a_{3} + a_{5} - . . . + a_{49}) / (a_{2} - a_{4} + a_{6} - . . . + a_{50})$ :
Given that $a_{1}, a_{2}, . . . a_{50}$ are in GP
Let $a$ be the first time and $r$ be the common ratio
$a_{1} = a a_{2} = a r a_{3} = a r^{2} a_{n} = a r^{n - 1} (a_{1} - a_{3} + a_{5} - . . . + a_{49}) / (a_{2} - a_{4} + a_{6} - . . . + a_{50}) = (a - a r^{2} + a r^{4} + . . . a r^{48}) / (a r - a r^{3} + a r^{5} + . . . a r^{49}) = (a - a r^{2} + a r^{4} + . . . a r^{48}) / r (a - a r^{2} + a r^{4} + . . . . a r^{48}) = 1 / r = a / a r = a 1 / a 2$
Hence, Option (C) is correct.

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Similar questions

a_{1}, a_{2}, . . . . . . . . . . . . ., a_{50}

G . P .

\frac{a_{1} - a_{3} + a_{5} - . . . . . + a_{49}}{a_{2} - a_{4} + a_{6} - . . . . . + a_{50}}

a_{1}, a_{2}, a_{3}, a_{4}, a_{5} a n d a_{6}

are consecutive odd natural numbers such that

a_{6} > a_{1}, a_{1} + a_{2} + a_{3} + a_{4} + a_{5} + a_{6} = p^{3} a n d a_{5} = 39

, then the value of

(a_{3} + a_{4} - 12 p)

a_{1}

a_{2}

a_{3}

a_{2}

a_{3}

a_{4}

a_{3}

a_{4}

a_{5}

a_{1}

a_{3}

a_{5}

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 . P .

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

a_{1} + a_{3} + a_{5} = - 12

a_{1} a_{2} a_{3} = 8

then the value of

a_{2} + a_{4} + a_{6}

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