Q.The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.

Concept: Geometric Progression (G.P.) and Arithmetic Progression (A.P.)

Let the three numbers in G.P. be $a$, $ar$, $ar^2$. Their sum is:

After subtracting 1, 7, 21 respectively, we get the A.P. terms:

For an A.P., the middle term is the average of the first and third:

Simplifying:

Now divide the sum equation by this result:

Cross-multiplying:

$1 + r + r^2 = 7(r^2 - 2r + 1) \quad \Rightarrow \quad 1 + r + r^2 = 7r^2 - 14r + 7$ …

Concept and Intuition

When a problem involves both a geometric progression and an arithmetic progression, the key is to translate each condition into equations. For a G.P., the natural choice is to let the three numbers be $a$, $ar$, and $ar^2$ — this automatically captures the constant ratio $r$ between consecutive terms.

The subtraction then creates a new sequence: $a-1$, $ar-7$, $ar^2-21$. For this to be an A.P., the middle term must be the arithmetic mean of the first and third terms. That single equation, combined with the sum condition, gives us two equations in two unknowns ($a$ and $r$). The rest is algebra — but careful algebra, because the numbers can appear in either order.

Step-by-Step Solution

1. Set up the G.P.

Let the three numbers in geometric progression be $a$, $ar$, and $ar^2$, where $a$ is the first term and $r$ is the common ratio.

2. Use the sum condition.

The sum is 56, so:

3. Translate the A.P. condition.

After subtracting 1, 7, and 21 respectively, we get the sequence:

For three numbers to be in arithmetic progression, the middle term must equal the average of the first and third:

4. Simplify the A.P. equation.

Bring all terms to one side:

Notice that $r^2 - 2r + 1 = (r-1)^2$, so:

5. Solve the system.

From Equation 2: $a = \frac{8}{(r-1)^2}$

Substitute into Equation 1:

$\frac{8}{(r-1)^2}(1 + r + r^2) = 56$ …