A number consists of three digits which are in GP the sum of the right hand and left hand digits exceeds twice the middle digits by 1 and the sum of the left hand and middle digits is two thirds of the sum of the middle and right hand digits. Find the number.

Explanation

To solve the problem step by step, let's denote the three digits of the number as \( A \), \( AR \), and \( AR^2 \), where \( A \) is the first digit and \( R \) is the common ratio of the geometric progression (GP).

### Step 1: Set up the equations based on the problem statement.

1. The digits are in GP:
- The digits are \( A \), \( AR \), and \( AR^2 \).

2. The first condition states that the sum of the right-hand digit and the left-hand digit exceeds twice the middle digit by 1:
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