Three numbers are in an increasing geometric progression with common ratio \( r \). If the middle number is doubled, then the new numbers are in an arithmetic progression with common difference \( d \). If the fourth term of GP is \( 3 r^2 \), then \( r^2 - d \) is equal to :
Show Hint
When \( a, b, c \) are in GP and \( a, 2b, c \) are in AP, the common ratio \( r \) always satisfies \( r^2 - 4r + 1 = 0 \). Memorizing this specific quadratic can save time.
Step 1: Understanding the Concept:
We use the properties of Geometric Progressions (\( a, ar, ar^2 \)) and Arithmetic Progressions (\( 2b = a + c \)) to solve for the variables \( a, r, \) and \( d \). Step 2: Detailed Explanation:
1. Let GP be \( a, ar, ar^2 \). Condition for increasing: \( r>1 \).
2. Doubling middle term: \( a, 2ar, ar^2 \) are in AP.
\[ 2(2ar) = a + ar^2 \implies 4r = 1 + r^2 \implies r^2 - 4r + 1 = 0 \]
Solving for \( r \): \( r = \frac{4 \pm \sqrt{16 - 4}}{2} = 2 \pm \sqrt{3} \).
Since \( r>1 \), \( r = 2 + \sqrt{3} \).
3. Fourth term of GP is \( ar^3 = 3r^2 \).
\[ ar = 3 \implies a = \frac{3}{r} = \frac{3}{2 + \sqrt{3}} = 3(2 - \sqrt{3}) \]
4. Calculate \( d \): In the AP \( a, 2ar, ar^2 \), the difference is \( d = 2ar - a \).
\[ d = 2(3) - 3(2 - \sqrt{3}) = 6 - 6 + 3\sqrt{3} = 3\sqrt{3} \]
5. Calculate \( r^2 - d \):
\[ r^2 = (2 + \sqrt{3})^2 = 4 + 3 + 4\sqrt{3} = 7 + 4\sqrt{3} \]
\[ r^2 - d = (7 + 4\sqrt{3}) - 3\sqrt{3} = 7 + \sqrt{3} \] Step 3: Final Answer:
The result is \( 7 + \sqrt{3} \).
