Step-by-step explanation:

Given:-

The numbers 54, 71, 75 and 99

To find:-

What should be subtracted from each of the numbers 54, 71, 75 and 99, so that the remainders are in CONTINUED proportion ?

Solution:-

Given numbers are 54,71,75 and 99

Let the number should be subtracted from each of them be X

Then they BECOMES

(54-X) ,(71-X) ,(75-X) and (99-X)

If they are in proportion then

(54-X) :(71-X) ::(75-X) : (99-X)

EXTREMES = (54-X) and (99-X)

Product of extremes = (54-X) × (99-X)

=>54(99-X) -X(99-X)

=>5346 -54X -99X +X^2

=>5346 -153 X +X^2

Product of extremes = X^2-153X+5346

Means = (71-X) and (75-X)

Product of means = (71-X) ×(75-X)

=>71(75-X)-X(75-X)

=>5325-71X-75X+X^2

=>5325 -146 X +X^2

Product of means = X^2-146X+5325

We know that

In proportion,

The product of extremes = The Product of means

=>X^2-153X+5346 = X^2-146X+5325

On cancelling X^2 both sides then

=>-153X+5346 = -146X +5325

=>5346 -5325 = -146X+153X

=>21 = 7X

=>7X = 21

=>X =21/7

=>X = 3

Therefore,The number = 3

Answer:-

The required number for the given problem = 3

Check:-

If X = 3 then the numbers become

54-3=51

71-3=68

75-3=72

99-3=96

Product of extremes = 51×96= 4896

Product of means = 68×72= 4896

They are in proportion.

Verified the given relation.

Used formulae:-

1. EQUALITY of ratios is called the Proportion.
2. In proportion,

The product of extremes = The Product of means