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We are given that:

- Three consecutive terms of an arithmetic progression (AP) have numbers subtracted from them, specifically 1, 4, and 3.
- The resulting numbers form a geometric progression (GP).
- The sum of the three terms of the AP is 21.

Let the three consecutive terms of the AP be:
\[
a - d, \quad a, \quad a + d
\]
where \(a\) is the middle term, and \(d\) is the common difference of the AP.

According to the problem:

- Subtracting 1 from \(a - d\), we get \((a - d) - 1\).
- Subtracting 4 from \(a\), we get \(a - 4\).
- Subtracting 3 from \(a + d\), we get \((a + d) - 3\).

These three resulting terms are in geometric progression (GP), so we can write the condition for a GP:
\[
\frac{(a - 4)}{(a - d - 1)} = \frac{(a + d - 3)}{(a - 4)}
\]
By cross-multiplying, we get:
\[
(a - 4)^2 = (a - d - 1)(a + d - 3)
\]
Now expand both sides:

### Left-hand side (LHS):
\[
(a - 4)^2 = a^2 - 8a + 16
\]

### Right-hand side (RHS):
\[
(a - d - 1)(a + d - 3) = a^2 + ad - 3a - d - ad + 3d - d - 3
\]
Simplify the RHS:
\[
= a^2 - 4a - 4d - 3
\]

Thus, we have the equation:
\[
a^2 - 8a + 16 = a^2 - 4a - 4d - 3
\]
Simplify by subtracting \(a^2\) from both sides:
\[
-8a + 16 = -4a - 4d - 3
\]
Now simplify further:
\[
-4a + 16 + 3 = -4d
\]
\[
-4a + 19 = -4d
\]
Dividing by \(-4\), we get:
\[
a - \frac{19}{4} = d
\]

We also know that the sum of the three terms of the AP is 21:
\[
(a - d) + a + (a + d) = 21
\]
This simplifies to:
\[
3a = 21
\]
So:
\[
a = 7
\]

Substitute \(a = 7\) into the equation \(d = a - \frac{19}{4}\):
\[
d = 7 - \frac{19}{4} = \frac{28}{4} - \frac{19}{4} = \frac{9}{4}
\]

Thus, the three terms of the AP are:
\[
a - d = 7 - \frac{9}{4} = \frac{28}{4} - \frac{9}{4} = \frac{19}{4}
\]
\[
a = 7
\]
\[
a + d = 7 + \frac{9}{4} = \frac{28}{4} + \frac{9}{4} = \frac{37}{4}
\]

Therefore, the three terms of the arithmetic progression are:
\[
\frac{19}{4}, \quad 7, \quad \frac{37}{4}
\]

Would you like further details or any additional questions?

Here are 5 related questions for further exploration:
1. What are the general properties of arithmetic and geometric progressions?
2. How can the method of cross-multiplication be used in solving ratio-based equations?
3. What happens if the common difference in an arithmetic progression is zero?
4. How can we derive the sum of any finite arithmetic progression?
5. Can geometric progressions always be derived from any arithmetic progression under specific conditions?

**Tip:** When dealing with both arithmetic and geometric sequences in a problem, always set up both sets of conditions carefully before solving for unknowns.

If the numbers 1, 4, and 3 are subtracted from three consecutive terms of an arithmetic progression, the resulting numbers are in geometric progression. Find the numbers if their sum is 21.

Learn how to solve a challenging problem involving arithmetic and geometric progressions, where three consecutive terms of an arithmetic progression have numbers subtracted, forming a geometric progression. The sum of the AP terms is 21. Suitable for advanced high school students in Grades 11-12.

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