Question

If $a_1,a_2,.............,a_{50}$ are in $G.P.$, then $\frac{a_1-a_3+a_5-.....+a_{49}}{a_2-a_4+a_6-.....+a_{50}}$

• A. $0$
• B. $1$
• C. $\frac{a_1}{a_2}$
• D. $\frac{a_{25}}{a_{24}}$

Answer 1

The correct option is C $\frac{a_1}{a_2}$

Given that $a_1,........,a_{50}$ are in G.P

(Let) $\to$ then $a_1=a$ & $r=$ ratio factor

then $a_2=ar,a_3=a{r}^2,.....,a_{50}=a{r}^{49}$

$\Rightarrow \frac{a_1-a_3+a_5.....+a_{49}}{a_2-a_4{a}_6....+5_{50}}$

$\Rightarrow \frac{a-a{r}^2+a{r}^2+a{r}^4......+a{r}^{48}}{ar-a{r}^3+a{r}^5......+a{r}^{49}}$

$=\frac{a(1-r^2+r^4........+r^{48})}{ar(1-r^2+r^4......+r^{48})}$

$=\frac{1}{r}=\frac{a_1}{a_2}=\frac{a}{ar}=\frac{1}{r}$

$[\frac{a_{2}4}{a_{2}5}=\frac{a{r}^{23}}{a{r}^{24}}=\frac{1}{r}]$

So, option (c) is correct.