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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 $\frac{a_1 - a_3 + a_5 - \dots + a_{49}}{a_2 - a_4 + a_6 - \dots + a_{50}} = \dots$

Similar Questions

$0.14189189189...$ can be expressed as a rational number.

If the roots of the equation $3x^3-26x^2+52x-24=0$ are in geometric progression,then the sum of two of its roots is

If $x > 1, y > 1, z > 1$ are in geometric progression,then in which progression are $\frac{1}{1 + \ln x}, \frac{1}{1 + \ln y}, \frac{1}{1 + \ln z}$?

If $A = 1 + r^z + r^{2z} + r^{3z} + \dots \infty$,then the value of $r$ is:

The third term of a geometric progression is equal to the square of the first term. If its second term is $8$,then its sixth term will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module