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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CAT 2020 Quant questions with Solutions