To solve this problem, let's denote the geometric progression (GP) as $a, ar, ar^2, ar^3, ar^4, \ldots$, where:

• $a$ is the first term, and
• $r$ is the common ratio.

Given:

• The 3rd term is 16: $ar^2 = 16$
• The 5th term is 256: $ar^4 = 256$

Step 1: Find the common ratio $r$

Dividing the equation for the 5th term by the equation for the 3rd term, we get: $\frac{ar^4}{ar^2} = \frac{256}{16}$ $r^2 = 16$ $r = \pm 4$

Step 2: Find the first term $a$

Using $r = 4$, substitute back into the equation for the 3rd term: $a \cdot 4^2 = 16$ $16a = 16$ $a = 1$

Alternatively, using $r = -4$: $a \cdot (-4)^2 = 16$ $16a = 16$ $a = 1$

So, the first term $a = 1$ for both cases.

Step 3: Find the 7th term of the series

The 7th term is given by: $a r^6 = 1 \cdot (\pm 4)^6 = 1 \cdot 4096 = 4096$

Step 4: Find the sum of the first 5 terms

The sum of the first $n$ terms of a geometric progression is given by: $S_n = a \frac{r^n - 1}{r - 1}, \text{ for } r \neq 1$ Substitute $a = 1$, $r = 4$, and $n = 5$: $S_5 = 1 \cdot \frac{4^5 - 1}{4 - 1} = \frac{1024 - 1}{3} = \frac{1023}{3} = 341$

For $r = -4$: $S_5 = 1 \cdot \frac{(-4)^5 - 1}{-4 - 1} = \frac{-1024 - 1}{-5} = \frac{-1025}{-5} = 205$

Summary

• First term, $a$ = 1
• Common ratio, $r$ = 4 or -4
• 7th term = 4096
• Sum of the first 5 terms = 341 (for $r = 4$) or 205 (for $r = -4$)

Would you like to go through any step in more detail, or have any questions?

Related Questions:

1. How do you determine the common ratio in a geometric progression?
2. What happens to a geometric progression if the common ratio is a fraction?
3. How does the formula for the sum of a geometric series change when $r = 1$?
4. What is the effect of a negative common ratio on the terms of a geometric sequence?
5. How do you find the sum to infinity of a geometric series?

Tip: The sum of an infinite geometric series converges only if the common ratio $|r| < 1$.