If a_1, a_2, dots, a_{50} are in a geometric | vedclass

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Question

(C) Let the geometric progression be $a, ar, ar^2, \dots, ar^{49}$.Then $a_n = ar^{n-1}$.The numerator is $a_1 - a_3 + a_5 - \dots + a_{49} = a - ar^2

Vedclass · Accepted Answer

$\frac{a_1}{a_2}$

Vedclass · Answer

(C) Let the geometric progression be $a, ar, ar^2, \dots, ar^{49}$.Then $a_n = ar^{n-1}$.The numerator is $a_1 - a_3 + a_5 - \dots + a_{49} = a - ar^2 + ar^4 - \dots + ar^{48}$.This is a geometric series with first term $A = a$,common ratio $R = -r^2$,and $n = 25$ terms.The sum is $\frac{a(1 - (-r^2)^{25})}{1 - (-r^2)} = \frac{a(1 + r^{50})}{1 + r^2}$.The denominator is $a_2 - a_4 + a_6 - \dots + a_{50} = ar - ar^3 + ar^5 - \dots + ar^{49}$.This is a geometric series with first term $A' = ar$,common ratio $R = -r^2$,and $n = 25$ terms.The sum is $\frac{ar(1 - (-r^2)^{25})}{1 - (-r^2)} = \frac{ar(1 + r^{50})}{1 + r^2}$.Dividing the numerator by the denominator,we get $\frac{a(1 + r^{50}) / (1 + r^2)}{ar(1 + r^{50}) / (1 + r^2)} = \frac{a}{ar} = \frac{1}{r} = \frac{a_1}{a_2}$.

If $a_1, a_2, \dots, a_{50}$ are in a geometric progression,then $\frac{a_1 - a_3 + a_5 - \dots + a_{49}}{a_2 - a_4 + a_6 - \dots + a_{50}} = \dots$

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