Question

the integer X, Y and Z each are perfect squares X greater than y greater than z greater than zero if x, y, z from an AP the smallest possible value of X is
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Answer 1

We are given 3 perfect square numbers X, Y and Z such that:

X > Y > Z > 0

These numbers are in AP i.e. Arithmetic Progression.

Arithmetic Progression is a series of numbers in which the next number is the previous number added to the common difference.

The next terms are found by adding a common difference 'd' to the previous numbers.

Here, we are given that Z, Y and X are in AP and X is the largest number.

Z is smallest.

As per the definition of common difference:

$X - Y = Y - Z \\\Rightarrow X + Z = 2 \times Y$

If we write the first 10 perfect squares greater than 0, these are:

1, 4, 9, 16, 25, 36, 49, 64, 81, 100

By looking at the above perfect squares, it is easily evident that the numbers 1, 25 and 49 are in AP.

Here, $X + Z = 50$

And $2 \times Y = 50$

i.e. Z is 1, Y is 25 and X is 49.

The smallest possible value of X is 49.