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CAT 2020 Slot 2 — QA Question 1

Geometric ProgressionEasy

Let the m-th and n-th terms of a geometric progression be 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

Answer & solution

  • A

    6

  • B

    2

  • C

    -4

  • -2

Solution

Easy

Divide the two given terms of the GP to eliminate the first term. This turns the data into a single equation in the integer ratio r n−m=16r^{\,n-m}=16, after which we test each integer rr and pick the combination giving the smallest r+n−mr+n-m.

1

Write the two terms. With first term aa and common ratio rr, the mm-th and nn-th terms are

Tm=a r m−1=34Tn=a r n−1=12\begin{aligned} &T_m = a\,r^{\,m-1} = \tfrac{3}{4}\\ &T_n = a\,r^{\,n-1} = 12 \end{aligned}
2

Divide to remove aa. Dividing TnT_n by TmT_m cancels the first term.

TnTm=a r n−1a r m−1=r n−m⇒ r n−m=123/4=16(from step 1)\begin{aligned} &\frac{T_n}{T_m} = \frac{a\,r^{\,n-1}}{a\,r^{\,m-1}} = r^{\,n-m}\\ &\Rightarrow\ r^{\,n-m} = \frac{12}{3/4} = 16 \quad\text{(from step 1)} \end{aligned}
3

List integer ratios. Since rr is an integer and n−m≥1n-m\ge 1 (because $m r=±2, n−m=4r=±4, n−m=2r=16, n−m=1\begin{aligned} &r=\pm 2,\ n-m=4\\ &r=\pm 4,\ n-m=2\\ &r=16,\ n-m=1 \end{aligned}

4

Minimise r+n−mr+n-m. Evaluate each case, taking the negative root of rr wherever allowed to make the sum small.

r=−2:  r+n−m=−2+4=2r=−4:  r+n−m=−4+2=−2r=16:  r+n−m=16+1=17\begin{aligned} &r=-2:\ \ r+n-m = -2+4 = 2\\ &r=-4:\ \ r+n-m = -4+2 = -2\\ &r=16:\ \ r+n-m = 16+1 = 17 \end{aligned}

The smallest value comes from r=−4, n−m=2r=-4,\ n-m=2.

r+n−m=−2r+n-m = -2
CAT 2020 Slot 2 QA Q1: Let the m-th and n-th terms of a geometric progression be 3/4 and 12, respectively, where m < n. If the common — Solution | TheCATExam