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: Set up equations based on given conditions

1. The sum of the terms is given as 126:
   \[
   a + ar + ar^2 + ar^3 = 126
   \]
   Factoring out \( a \) from the left-hand side:
   \[
  

if the sum and product of four consecutive terms of GP are 126 and 1296 respectively then the sum of common ratios of all such GPs is?

This problem involves solving for the common ratios of a geometric progression (GP) with four consecutive terms, where the sum is 126 and the product is 1296. Learn how to set up and solve the equations step-by-step, covering important GP properties. Suitable for Grades 11-12.

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