three number `a,b,c` are in `GP` such that : (i) `a+b+c=70` (ii) `4a,5b,4c` are in `AP` if `G=max{a,b,c}` and `L=min{a,b,c}` than `[G/L]` is equal is :

Options

A. 2
B. 3
C. 5
D. 4

Answer

To solve the problem step by step, we will follow the given conditions and derive the values of \( a \), \( b \), and \( c \) systematically.

### Step 1: Express \( a \), \( b \), and \( c \) in terms of a common ratio \( r \)

Since \( a \), \( b \), and \( c \) are in geometric progression (GP), we can express them as:
- \( a = a \)
- \( b = ar \)
- \( c = ar^2 \)

### Step 2: Use the sum condition

From the problem, we know that:
\[ a + b + c = 70 \]
Substituting the expressions for \( b \) and \( c \):
\[ a + ar + ar^2 = 70 \]
Factoring out \( a \):
\[ a(1 + r + r^2) = 70 \]

### Step 3: Use the condition for \( 4a, 5b, 4c \) being in arithmetic progression (AP)

The numbers \( 4a, 5b, 4c \) are in AP, which means:
\[ 2(5b) = 4a + 4c \]
Substituting \( b \) and \( c \):
\[ 10ar = 4a + 4(ar^2) \]
This simplifies to:
\[ 10ar = 4a + 4ar^2 \]
Rearranging gives:
\[ 10ar - 4ar^2 - 4a = 0 \]
Factoring out \( 2a \):
\[ 2a(5r - 2r^2 - 2) = 0 \]
Since \( a \neq 0 \), we can divide by \( 2a \):
\[ 5r - 2r^2 - 2 = 0 \]

### Step 4: Solve the quadratic equation

Rearranging gives:
\[ 2r^2 - 5r + 2 = 0 \]
Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ r = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot 2}}{2 \cdot 2} \]
\[ r = \frac{5 \pm \sqrt{25 - 16}}{4} \]
\[ r = \frac{5 \pm 3}{4} \]
This gives us two possible values for \( r \):
1. \( r = 2 \)
2. \( r = \frac{1}{2} \)

### Step 5: Find \( a \) for each value of \( r \)

**Case 1: \( r = 2 \)**
\[ a(1 + 2 + 4) = 70 \implies a \cdot 7 = 70 \implies a = 10 \]
Thus:
- \( b = ar = 10 \cdot 2 = 20 \)
- \( c = ar^2 = 10 \cdot 4 = 40 \)

**Case 2: \( r = \frac{1}{2} \)**
\[ a(1 + \frac{1}{2} + \frac{1}{4}) = 70 \implies a \cdot \frac{7}{4} = 70 \implies a = 40 \]
Thus:
- \( b = ar = 40 \cdot \frac{1}{2} = 20 \)
- \( c = ar^2 = 40 \cdot \frac{1}{4} = 10 \)

### Step 6: Calculate \( G \) and \( L \)

In both cases:
- The maximum \( G = 40 \)
- The minimum \( L = 10 \)

### Step 7: Calculate \( \frac{G}{L} \)

\[ \frac{G}{L} = \frac{40}{10} = 4 \]

### Final Answer

Thus, the value of \( \frac{G}{L} \) is \( \boxed{4} \).

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