Question

Let $A_r,r=1,2,3,.....$ be points on the number line such that $O{A}_1,O{A}_2,O{A}_3,......$ are in G.P., where $O$ is the origin and the common ratio of the G.P. be a positive proper fraction. Let $M_r$, be the middle point of the line segmentn$A_r{A}_{r+1}$. Then the value of ${\sum }_{r=1}^{\infty }O{M}_r$ is equal to

• A. $\frac{O{A}_1(O{A}_1-O{A}_2)}{2(O{A}_1+O{A}_2)}$
• B. $\frac{O{A}_1(O{A}_1+O{A}_2)}{2(O{A}_1-O{A}_2)}$
• C. $\frac{O{A}_1}{2(O{A}_1-O{A}_2)}$
• D. $\infty$

Answer 1

Answer: (B)

$\frac{O{A}_1(O{A}_1+O{A}_2)}{2(O{A}_1-O{A}_2)}$

We have,
$O{A}_1=a_{1}d$ ($d$ is the common ratio)
$O{A}_2=a_1{d}^2$
Similarly we can write the general term as :
$O{A}_k=a_1{d}^k$
Now, the midpoint of $A_r{A}_{r+1}$ can be written as $\frac{a_1(d^{r+1}+d^r)}{2}$

$O{M}_r=\frac{a_1(d^{r+1}+d^r)}{2}$
${\sum }_{r=1}^{r=\infty }O{M}_r=\sum _{r=1}^{r=\infty }\frac{a_1(d^r+d^{r+1})}{2}$
This is an infinite geometric series, with $a=\frac{a_{1}d(1+d)}{2}$, $r=d$
Hence,
Sum $=\frac{a_1\left(d\right)(1+d)}{2(1-d)}$

Sum $=\frac{a_1(d^2+d)}{2(1-d)}$

Sum $=\frac{a_{1}d(d^2+d)}{2(d-d^2)}$

Sum $=\frac{O{A}_1(O{A}_2+O{A}_1)}{2(O{A}_1-O{A}_2)}$