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Solve question number 5 of sequence and series in a easy way

To solve Question 5 of Sequence and Series, let's break it down step-by-step.

### Problem:
We are given three numbers in a Geometric Progression (GP) whose product is 56. These numbers, when subtracted by 1, 7, and 21 respectively, form an Arithmetic Progression (AP).

We need to find the three numbers.

### Let the three numbers be:
\[ a, ar, ar^2 \]

Here, \( a \) is the first term, \( r \) is the common ratio, and \( ar^2 \) is the third term in the GP.

### Step 1: Product of the three numbers
According to the problem, the product of the three numbers in the GP is 56. So, we have:
\[
a \cdot ar \cdot ar^2 = 56
\]
This simplifies to:
\[
a^3 r^3 = 56 \quad \text{or} \quad (a \cdot r)^3 = 56
\]
So,
\[
a \cdot r = \sqrt[3]{56}
\]
\[
a \cdot r = 3.83 \quad \text{(approximately)}
\]

### Step 2: Conditions for Arithmetic Progression (AP)
If we subtract 1, 7, and 21 from the terms of the GP, the resulting numbers form an AP. This means:
\[
a - 1, ar - 7, ar^2 - 21 \quad \text{are in AP}.
\]
For numbers to form an AP, the middle term should be the average of the other two. So:
\[
ar - 7 = \frac{(a - 1) + (ar^2 - 21)}{2}
\]
Simplify the right-hand side:
\[
ar - 7 = \frac{a - 1 + ar^2 - 21}{2}
\]
\[
ar - 7 = \frac{a + ar^2 - 22}{2}
\]
Multiply both sides by 2 to eliminate the fraction:
\[
2(ar - 7) = a + ar^2 - 22
\]
\[
2ar - 14 = a + ar^2 - 22
\]
Rearrange the terms:
\[
2ar - ar^2 = a - 8
\]
This is the equation that we can use to solve for \( a \) and \( r \).

### Step 3: Solve the equations
Now, we have two equations:
1. \( a^3 r^3 = 56 \) (from the product of terms)
2. \( 2ar - ar^2 = a - 8 \) (from the AP condition)

You can either solve these equations algebraically or use a numerical method to find \( a \) and \( r \).

Let me know if you'd like further assistance with these steps or solving this system!

The sum of three numbers in a GP is 56. If we subtract 1, 7, and 21 respectively from these numbers, they will form an AP. Find the numbers.

Step-by-step solution to finding three numbers in a GP whose sum is 56, and when subtracted by 1, 7, and 21, form an AP. Learn to use GP and AP properties and solve algebraic equations. Suitable for students in Grades 9-11.

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