CAT 2021 Slot 2 · Quant · Question 10

In an AP the middle term is the average, so x+z=2y and hence x+y+z=3y.

The condition becomes xyz=15y, and since y>0 we may cancel it: xz=15.

Write d=y-x, the common difference, so x=y-d and z=y+d. Then xz={y}^{2}-{d}^{2}=15, that is {y}^{2}=15+{d}^{2}.

We need y to be a positive integer with d>2. Trying d=7 gives {y}^{2}=64,

so y=8, x=1 and z=15.

Check: xz=15, and xyz=1\times 8\times 15=120

while 5\left( x+y+z \right)=5\times 24=120. So z-x=2d=14.