For an increasing G.P. `a_(1), a_(2), a_(3),.....a_(n), " If " a_(6) = 4a_(4), a_(9) - a_(7) = 192`, then the value of `sum_(l=1)^(oo) (1)/(a_(i))` is

Answer

To solve the problem, we need to find the value of the infinite series \( \sum_{i=1}^{\infty} \frac{1}{a_i} \) given the conditions for the terms of an increasing geometric progression (G.P.).

### Step 1: Define the terms of the G.P.
Let the first term of the G.P. be \( a \) and the common ratio be \( r \). Then the terms can be expressed as:
- \( a_1 = a \)
- \( a_2 = ar \)
- \( a_3 = ar^2 \)
- \( a_4 = ar^3 \)
- \( a_5 = ar^4 \)
- \( a_6 = ar^5 \)
- \( a_7 = ar^6 \)
- \( a_8 = ar^7 \)
- \( a_9 = ar^8 \)

### Step 2: Use the first condition \( a_6 = 4a_4 \)
From the problem, we know:
\[ a_6 = 4a_4 \]
Substituting the expressions for \( a_6 \) and \( a_4 \):
\[ ar^5 = 4(ar^3) \]
Dividing both sides by \( a \) (assuming \( a \neq 0 \)):
\[ r^5 = 4r^3 \]
Dividing both sides by \( r^3 \) (assuming \( r \neq 0 \)):
\[ r^2 = 4 \]
Taking the square root:
\[ r = 2 \quad (\text{since the G.P. is increasing, we take the positive root}) \]

### Step 3: Use the second condition \( a_9 - a_7 = 192 \)
Now, we use the second condition:
\[ a_9 - a_7 = 192 \]
Substituting the expressions for \( a_9 \) and \( a_7 \):
\[ ar^8 - ar^6 = 192 \]
Factoring out \( ar^6 \):
\[ ar^6(r^2 - 1) = 192 \]
Substituting \( r = 2 \):
\[ ar^6(2^2 - 1) = 192 \]
This simplifies to:
\[ ar^6(4 - 1) = 192 \]
\[ ar^6 \cdot 3 = 192 \]
\[ ar^6 = \frac{192}{3} = 64 \]

### Step 4: Calculate \( a \)
Now substituting \( r = 2 \):
\[ a(2^6) = 64 \]
\[ a \cdot 64 = 64 \]
Thus:
\[ a = 1 \]

### Step 5: Find the sum \( \sum_{i=1}^{\infty} \frac{1}{a_i} \)
Now we can find the sum:
\[ \sum_{i=1}^{\infty} \frac{1}{a_i} = \sum_{i=1}^{\infty} \frac{1}{ar^{i-1}} = \sum_{i=1}^{\infty} \frac{1}{1 \cdot 2^{i-1}} = \sum_{i=1}^{\infty} \frac{1}{2^{i-1}} \]
This is a geometric series with first term \( 1 \) and common ratio \( \frac{1}{2} \):
\[ \sum_{i=1}^{\infty} \frac{1}{2^{i-1}} = \frac{1}{1 - \frac{1}{2}} = 2 \]

### Final Answer
The value of \( \sum_{i=1}^{\infty} \frac{1}{a_i} \) is \( \boxed{2} \).