Q. If $a_{1},a_{2},\ldots.a_{50}$ are in GP, then $\frac{a_{1} - a_{3} + a_{5} - ... + a_{49}}{a_{2} - a_{4} + a_{6} - ... + a_{50}}$ is equal to
A) 0
B) 1
C) $\frac{a_{1}}{a_{2}}$
D) $\frac{a_{25}}{a_{24}}$
Explanation
(c)
Let $a_{2} = ra_{1},a_{3} = r^{2}a_{1},\ldots,$so on
$\therefore\ \frac{a_{1} - a_{3} + a_{5} - ... + a_{49}}{a_{2} - a_{4} + a_{6} - ... + a_{50}}$
$= \frac{a_{1} - r^{2}a_{1} + r^{4}\ a_{1} - ... + r^{48}a_{1}}{a_{1}r - r^{3}a_{1} + r^{5}a_{1} - ... + r^{49}a_{1}}$
$= \frac{a_{1}(1 - r^{2} + r^{4} - ... + r^{48})}{a_{1}r(1 - r^{2} + r^{4} - ... + r^{48})}$
$= \frac{1}{r} = \frac{a_{1}}{a_{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.