To solve this problem, we need to use the properties of a geometric progression (GP).
Let the first term of the GP be \( a \), and the common ratio be \( r \).
The \( n \)-th term of a GP can be expressed as: \[ T_n = a r^{(n-1)} \]
From the problem, we're given:
- The 3rd term \( T_3 = 54 \): \[ T_3 = a r^{2} = 54 \]
- The 9th term \( T_9 = 39366 \): \[ T_9 = a r^{8} = 39366 \]
We can set up a system of equations based on these two terms:
- \( a r^2 = 54 \) (Equation 1)
- \( a r^8 = 39366 \) (Equation 2)
Now, let's solve these equations. We can express \( a \) from Equation 1: \[ a = \frac{54}{r^2} \]
Substituting \( a \) into Equation 2: \[ \frac{54}{r^2} r^8 = 39366 \] \[ 54 r^6 = 39366 \]
Now, we can solve for \( r^6 \): \[ r^6 = \frac{39366}{54} \] \[ r^6 = 729 \]
Taking the sixth root of both sides: \[ r = 3 \]
Now substituting \( r = 3 \) back into Equation 1 to find \( a \): \[ a (3^2) = 54 \] \[ a \cdot 9 = 54 \] \[ a = \frac{54}{9} = 6 \]
So we have:
- \( a = 6 \)
- \( r = 3 \)
Now we can find the first few terms of the GP:
- \( T_1 = a = 6 \)
- \( T_2 = 6 \cdot 3 = 18 \)
- \( T_3 = 54 \) (given)
- \( T_4 = 54 \cdot 3 = 162 \)
- \( T_5 = 162 \cdot 3 = 486 \)
- \( T_6 = 486 \cdot 3 = 1458 \)
- \( T_7 = 1458 \cdot 3 = 4374 \)
- \( T_8 = 4374 \cdot 3 = 13122 \)
- \( T_9 = 39366 \) (given)
1. Sixth Term:
The 6th term \( T_6 \) is: \[ T_6 = 1458 \]
2. Sum of the 4th and 7th Term:
The 4th and 7th terms are: \[ T_4 = 162 \] \[ T_7 = 4374 \] The sum is: \[ T_4 + T_7 = 162 + 4374 = 4536 \]
3. Product of the 2nd and 5th Term:
The 2nd and 5th terms are: \[ T_2 = 18 \] \[ T_5 = 486 \] The product is: \[ T_2 \cdot T_5 = 18 \cdot 486 = 8748 \]
Final Answers:
- Sixth term \( T_6 = 1458 \)
- Sum of the 4th and 7th term \( = 4536 \)
- Product of the 2nd and 5th term \( = 8748 \)