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# CAT 2020 Quant Question [Slot 2] with Solution 05

Question

Let the m-th and n-th terms of a geometric progression be $\frac{3}{4}$ and 12 , respectively, where m<n. If the common ratio of the progression is an integer r, then the smallest possible value of r + n - m is

1. -2
2. 2
3. 6
4. -4
Option: 1
Solution:

${T_n} = 12$

${T_m} = 3/4$

$\frac{{{T_n}}}{{{T_m}}} = \frac{{a{r^{n - 1}}}}{{a{r^{m - 1}}}} = \frac{{12}}{{\frac{3}{4}}}$

${r^{n - m}} = 16 = {( \pm 2)^4} = {( \pm 4)^2}$

To get the minimum value for $r + n - m,r$ should be minimum.

$\therefore r = - 4$

$n - m = 2$

$\therefore$ Required answer =-2

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