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CAT 2018 Slot 1 — QA Question 34

Geometric ProgressionEasy

Let x, y, z be three positive real numbers in a geometric progression such that x < y < z. If 5x, 16y, and 12z are in an arithmetic progression then the common ratio of the geometric progression is

Answer & solution

Correct answer: 5/2

  • A

    1/6

  • 5/2

  • C

    3/6

  • D

    3/2

Solution

Easy

Write the three GP terms using the middle term yy and ratio kk. Apply the AP condition (middle term is the average of its neighbours) to get a quadratic in kk, then keep the root consistent with $x

1

Express the GP terms. Let the common ratio be kk, so x=ykx=\dfrac{y}{k} and z=ykz=yk.

5x=5yk,16y,12z=12yk\begin{aligned} &5x = \frac{5y}{k},\qquad 16y,\qquad 12z = 12yk \end{aligned}
2

Apply the AP condition. For an AP, twice the middle term equals the sum of the outer terms. Divide through by yy.

2(16y)=5yk+12yk⇒ 32=5k+12k(divide by y)⇒ 12k2−32k+5=0(×k)\begin{aligned} &2(16y) = \frac{5y}{k} + 12yk\\ &\Rightarrow\ 32 = \frac{5}{k} + 12k \quad\text{(divide by }y)\\ &\Rightarrow\ 12k^2 - 32k + 5 = 0 \quad\text{(}\times k\text{)} \end{aligned}
3

Solve the quadratic and pick the valid root.

12k2−32k+5=0⇒ (2k−5)(6k−1)=0⇒ k=52 or k=16\begin{aligned} &12k^2 - 32k + 5 = 0\\ &\Rightarrow\ (2k-5)(6k-1)=0\\ &\Rightarrow\ k = \tfrac{5}{2} \ \text{or}\ k = \tfrac{1}{6} \end{aligned}

Since $x1.Thus. Thusk=\tfrac52$.

k=52k = \tfrac{5}{2}

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CAT 2018 Slot 1 QA Q34: Let x, y, z be three positive real numbers in a geometric progression such that x < y < z. If 5x, 16y, and 12z — Solution | TheCATExam