Question

The 1st and 20th term of a GP are 512 and \( \frac{1}{1024} \), respectively. Find the common ratio.

Hint:

The formula for the nth term of a GP, \( T_n = a r^{n-1} \), helps us find the common ratio when given terms in the progression. In this case, we used the relation between the 1st and 20th terms to solve for \( r \).

Answer 1

In a geometric progression (GP), the nth term is given by the formula: \[ T_n = a r^{n-1} \] where:
- \( T_n \) is the nth term,
- \( a \) is the first term,
- \( r \) is the common ratio, and
- \( n \) is the term number.
We are given:
- The 1st term \( T_1 = 512 \),
- The 20th term \( T_{20} = \frac{1}{1024} \).
Step 1: Use the formula for the nth term for the 1st term and 20th term.
For the 1st term, we have: \[ T_1 = a r^{0} = a = 512 \] So, \( a = 512 \). For the 20th term, we have: \[ T_{20} = a r^{19} = \frac{1}{1024} \] Substitute \( a = 512 \) into this equation: \[ 512 r^{19} = \frac{1}{1024} \]
Step 2: Solve for \( r \).
Now, solve for \( r \): \[ r^{19} = \frac{1}{1024} \times \frac{1}{512} = \frac{1}{1024 \times 512} = \frac{1}{2^{10} \times 2^9} = \frac{1}{2^{19}} \] Thus, we get: \[ r^{19} = \frac{1}{2^{19}} \] Taking the 19th root of both sides: \[ r = \frac{1}{2} \] Thus, the common ratio is: \[ \boxed{r = \frac{1}{2}} \]