Question

The sum of the first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the 1st term, the common ratio, and the sum to \( n \) terms of the G.P.

Hint:

To find the sum of the terms in a G.P., use the formula \( S_n = a \frac{1 - r^n}{1 - r} \) for \( r \neq 1 \). When the sum of terms is given, create a system of equations to find the first term and common ratio.

Answer 1

Step 1: Use the formula for the sum of the first \( n \) terms of a G.P.
The sum of the first \( n \) terms of a geometric progression is given by: \[ S_n = a \frac{1 - r^n}{1 - r} \quad (\text{for } r \neq 1) \] where \( a \) is the first term, and \( r \) is the common ratio.
Step 2: Sum of the first three terms.
We are given that the sum of the first three terms is 16: \[ a + ar + ar^2 = 16 \] \[ a(1 + r + r^2) = 16 \]
Step 3: Sum of the next three terms.
We are also given that the sum of the next three terms is 128: \[ ar^3 + ar^4 + ar^5 = 128 \] \[ a r^3 (1 + r + r^2) = 128 \] Since \( 1 + r + r^2 \) is common in both equations, we can divide the second equation by the first equation: \[ \frac{a r^3 (1 + r + r^2)}{a(1 + r + r^2)} = \frac{128}{16} \] \[ r^3 = 8 \] \[ r = 2 \]
Step 4: Find the first term.
Substitute \( r = 2 \) into the first equation: \[ a(1 + 2 + 2^2) = 16 \] \[ a(1 + 2 + 4) = 16 \] \[ a(7) = 16 \] \[ a = \frac{16}{7} \]
Step 5: Find the sum to \( n \) terms.
The sum to \( n \) terms of a G.P. is given by: \[ S_n = a \frac{1 - r^n}{1 - r} \] Substitute \( a = \frac{16}{7} \) and \( r = 2 \): \[ S_n = \frac{16}{7} \frac{1 - 2^n}{1 - 2} = \frac{16}{7} \times (2^n - 1) \]
Step 6: Conclusion.
The first term \( a = \frac{16}{7} \), the common ratio \( r = 2 \), and the sum to \( n \) terms is: \[ S_n = \frac{16}{7} (2^n - 1) \]