Concept:

If a, b, c are in GP then \( \dfrac{b}{a} = \dfrac{c}{b} \)

Calculations:

Consider, three positive numbers \( a - d \), \( a \), and \( a + d \) are in AP.

These three positive numbers have a sum of 21:

Hence, \( a - d + a + a + d = 21 \)

Therefore, \( a = 7 \).

Hence, the three positive numbers are \( 7 - d \), \( 7 \), and \( 7 + d \).

It is given that if 2, 2, and 14 are added to them respectively, the resulting numbers are in geometric progression.

Thus, \( 9 - d \), \( 9 \), and \( 21 + d \) are in GP.

We can write the equation as:

\( \dfrac{9}{9 - d} = \dfrac{21 + d}{9} \)

Hence, \( 81 = (21 + d)(9 - d) \)

Which simplifies to:

\( d^2 + 12d - 108 = 0 \)

Factoring the quadratic equation:

\( (d + 18)(d - 6) = 0 \)

Therefore, \( d = -18 \) or \( d = 6 \).

When \( d = -18 \), the three numbers are 25, 7, and -11.

When \( d = 6 \), the three numbers are 1, 7, and 13.