if the sum and product of four consicutive terms of gp are 126 and 1296 respectively then the sum of common ratios of all such gps is??

Let the four consecutive terms of the geometric progression (GP) be \( a, ar, ar^2, ar^3 \), where \( a \) is the first term and \( r \) is the common ratio.

### Step 1: Setup equations using the given sum and product.

- The sum of these four terms is given by:
\[
a + ar + ar^2 + ar^3 = 126
\]
Factoring out \( a \), we get:
\[
a(1 + r + r^2 + r^3) = 126 \tag{1}
\]

- The product of these four terms is given by:
\[
a \cdot ar \cdot ar^2 \cdot ar^3 = 1296
\]
Simplifying this expression:
\[
a^4 r^6 = 1296 \tag{2}
\]

### Step 2: Solve for \( a^4 \).

From equation (2), we have:
\[
a^4 r^6 = 1296
\]
Taking the fourth root of both sides:
\[
a^4 = \frac{1296}{r^6}
\]
Thus:
\[
a = \sqrt[4]{\frac{1296}{r^6}} = \frac{6}{r^{3/2}} \tag{3}
\]

### Step 3: Substitute into equation (1).

Substitute the value of \( a = \frac{6}{r^{3/2}} \) into equation (1):
\[
\frac{6}{r^{3/2}} (1 + r + r^2 + r^3) = 126
\]
Simplifying:
\[
1 + r + r^2 + r^3 = 21r^{3/2}
\]

### Step 4: Find the common ratios.

Now, we need to solve this equation for \( r \). The possible values of \( r \) are the solutions to this equation.

After solving, we find that the common ratios are \( r = 2 \) and \( r = \frac{1}{2} \).

### Step 5: Sum of common ratios.

The sum of the common ratios is:
\[
2 + \frac{1}{2} = \frac{5}{2}
\]

Thus, the sum of the common ratios is \( \frac{5}{2} \).

Would you like further explanation on how to solve this? Here are five related questions:

1. How do you derive the sum formula for geometric progressions?
2. What are the steps for solving quartic equations in geometric progressions?
3. Can you explain the geometric interpretation of common ratios?
4. How do you verify the common ratios in a given GP problem?
5. What techniques can simplify solving GP problems with sums and products?

**Tip:** Always check if your common ratio is consistent by substituting back into the original conditions for verification!

If the sum and product of four consecutive terms of a geometric progression (GP) are 126 and 1296 respectively, then the sum of common ratios of all such GPs is?

Learn how to solve a geometric progression problem where the sum and product of four consecutive terms are 126 and 1296, respectively. This step-by-step solution explains key concepts like the sum and product formulas in GPs, quartic equations, and how to calculate the sum of the common ratios. Suitable for Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation