Three distinct numbers `a_1 , a_2, a_3` are in increasing G.P. `a_1^2 + a_2^2 + a_3^2 = 364`and `a_1 + a_2 + a_3 = 26` then the value of `a_10` if `a_n` is the `n^(th)` term of the given G.P. is:

Options

A. `2.3^9`
B. `3^9`
C. `2.3^(10)`
D. `3^(12)`

Explanation

To solve the problem, we need to find the 10th term of a geometric progression (G.P.) given the conditions about the first three terms. Let's denote the three terms in the G.P. as \( a_1, a_2, a_3 \). Since they are in G.P., we can express them as follows:

1. Let \( a_1 = \frac{a}{r} \)
2. Let \( a_2 = a \)
3. Let \( a_3 = ar \)

where \( a \) is the middle term and \( r \) is the common ratio.

### Step 1: Set up the equations

From the problem, we have two equations:

1. \( a_1 + a_2 + a_3 = 26 \)
\[ \frac{a}{r} + a + ar = 26 \]
Multiplying through by \( r \) to eliminate the fraction gives:
\[ a + ar + a r^2 = 26r \]
This simplifies to:
\[ a(1 + r + r^2) = 26r \quad \text{(Equation 1)} \]

2. \( a_1^2 + a_2^2 + a_3^2 = 364 \)
\[ \left(\frac{a}{r}\right)^2 + a^2 + (ar)^2 = 364 \]
This simplifies to:
\[ \frac{a^2}{r^2} + a^2 + a^2r^2 = 364 \]
Multiplying through by \( r^2 \) gives:
\[ a^2 + a^2r^2 + a^2r^4 = 364r^2 \]
This simplifies to:
\[ a^2(1 + r^2 + r^4) = 364r^2 \quad \text{(Equation 2)} \]

### Step 2: Solve for \( a \) and \( r \)

From Equation 1, we can express \( a \):
\[ a = \frac{26r}{1 + r + r^2} \]

Substituting this expression for \( a \) into Equation 2 gives:
\[ \left(\frac{26r}{1 + r + r^2}\right)^2(1 + r^2 + r^4) = 364r^2 \]

Expanding this leads to:
\[ \frac{676r^2(1 + r^2 + r^4)}{(1 + r + r^2)^2} = 364r^2 \]

Cancelling \( r^2 \) (assuming \( r \neq 0 \)):
\[ \frac{676(1 + r^2 + r^4)}{(1 + r + r^2)^2} = 364 \]

Cross-multiplying gives:
\[ 676(1 + r^2 + r^4) = 364(1 + r + r^2)^2 \]

### Step 3: Solve the resulting polynomial

This equation can be simplified and solved for \( r \). After solving, we find \( r = 3 \) (since we are looking for an increasing G.P.).

### Step 4: Find \( a \)

Substituting \( r = 3 \) back into Equation 1 to find \( a \):
\[ a = \frac{26 \cdot 3}{1 + 3 + 9} = \frac{78}{13} = 6 \]

### Step 5: Find the 10th term \( a_{10} \)

The nth term of a G.P. is given by:
\[ a_n = a \cdot r^{n-1} \]
Thus, for \( n = 10 \):
\[ a_{10} = 6 \cdot 3^{10-1} = 6 \cdot 3^9 \]

Calculating \( 3^9 \):
\[ 3^9 = 19683 \]
Thus:
\[ a_{10} = 6 \cdot 19683 = 118098 \]

### Final Answer:
The value of \( a_{10} \) is \( 118098 \).