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CAT 2021 Slot 2 — QA Question 19

Arithmetic ProgressionEasy

Three positive integers x, y and z are in arithmetic progression. If y − x > 2 and xyz = 5(x + y + z), then z − x equals 

Answer & solution

  • A

    12

  • B

    8

  • 14

  • D

    10

Solution

Easy

Since x,y,zx,y,z are in AP, write them symmetrically as a−d, a, a+da-d,\,a,\,a+d. The condition y−x>2y-x>2 means d>2d>2. Substituting into xyz=5(x+y+z)xyz=5(x+y+z) collapses to (a−d)(a+d)=15(a-d)(a+d)=15; factor 1515 into two positive integers and pick the factorisation consistent with d>2d>2.

1

Set up symmetric terms. Let x=a−d, y=a, z=a+dx=a-d,\ y=a,\ z=a+d with positive integers. Then x+y+z=3ax+y+z=3a and y−x=d>2y-x=d>2.

xyz=5(x+y+z)⇒ (a−d) a (a+d)=5⋅3a\begin{aligned} &xyz=5(x+y+z)\\ &\Rightarrow\ (a-d)\,a\,(a+d)=5\cdot 3a \end{aligned}
2

Cancel aa and factor. Since a>0a>0, divide by aa.

(a−d)(a+d)=15⇒ a2−d2=15\begin{aligned} &(a-d)(a+d)=15\\ &\Rightarrow\ a^{2}-d^{2}=15 \end{aligned}

Write 1515 as a product of two positive integers (a−d)(a+d)(a-d)(a+d): either 1×151\times 15 or 3×53\times 5.

3

Solve both factorisations and apply d>2d>2.

a−d=1, a+d=15 ⇒ (a,d)=(8,7)a−d=3, a+d=5 ⇒ (a,d)=(4,1)\begin{aligned} &a-d=1,\ a+d=15\ \Rightarrow\ (a,d)=(8,7)\\ &a-d=3,\ a+d=5\ \Rightarrow\ (a,d)=(4,1) \end{aligned}

d>2d>2 rejects (4,1)(4,1), leaving (a,d)=(8,7)(a,d)=(8,7), i.e. x,y,z=1,8,15x,y,z=1,8,15.

z−x=(a+d)−(a−d)=2d=14\begin{aligned} &z-x=(a+d)-(a-d)=2d=14 \end{aligned}
z−x=14z-x=14
CAT 2021 Slot 2 QA Q19: Three positive integers x, y and z are in arithmetic progression. If y − x > 2 and xyz = 5(x + y + z), t — Solution | TheCATExam