Q31 / 100 · BITSAT 2024 (Memory Based Paper 2) · Mathematics

If $A=1+r^a+r^{2 a}+r^{3 a}+\ldots \infty$ and $\mathrm{B}=1+\mathrm{r}^{\mathrm{b}}+\mathrm{r}^{2 \mathrm{~b}}+\mathrm{r}^{3 \mathrm{~b}}+\ldots \infty$, then $\frac{\mathrm{a}}{\mathrm{b}}$ is equal to

  • A
    $\log _B(A)$
  • B
    $\log _{1-B}(1-A)$
  • Correct answer
    C
    $\log _{\frac{B-1}{B}}\left(\frac{A-1}{A}\right)$
  • D
    None of these

Solution

$A=\frac{1}{1-r^a} \Rightarrow 1-r^a=\frac{1}{A} \Rightarrow r^a=1-\frac{1}{A}=\frac{A-1}{A}$ $B=\frac{1}{1-r^b} \Rightarrow 1-r^b=\frac{1}{B} \Rightarrow r^b=1-\frac{1}{B}=\frac{B-1}{B}$ $\therefore a \log r=\log \left(\frac{A-1}{A}\right)$ and $b \log r=\log \left(\frac{B-1}{B}\right)$ $\therefore \frac{\mathrm{a}}{\mathrm{b}}=\frac{\log \left(\frac{\mathrm{A}-1}{\mathrm{~A}}\right)}{\log \left(\frac{\mathrm{B}-1}{\mathrm{~B}}\right)}=\log _{\frac{\mathrm{B}-1}{\mathrm{~B}}}\left(\frac{\mathrm{~A}-1}{\mathrm{~A}}\right)$

Hence, the correct answer is option: 'C'.

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