The problem asks for the sum of the 5th and 7th terms of a geometric progression (GP) where the 3rd term is 63, and the 5th term is 567. Let us solve it step by step.

Step 1: Formula for the nth term of a GP

The general formula for the $n$-th term of a GP is: $T_n = a \cdot r^{n-1}$ Where:

• $a$ = the first term
• $r$ = the common ratio
• $n$ = the term position

Step 2: Write the equations for given terms

• For the 3rd term ($T_3$): $T_3 = a \cdot r^{3-1} = a \cdot r^2 = 63$
• For the 5th term ($T_5$): $T_5 = a \cdot r^{5-1} = a \cdot r^4 = 567$

Step 3: Solve for $r$

Divide the equation for $T_5$ by the equation for $T_3$: $\frac{T_5}{T_3} = \frac{a \cdot r^4}{a \cdot r^2} = r^{4-2} = r^2$ $\frac{567}{63} = r^2$ r^2 = 9 \implies r = 3 \; \text{(since r > 0 in this GP)}

Step 4: Solve for $a$

Substitute $r = 3$ into the equation for $T_3$: $a \cdot r^2 = 63$ $a \cdot 3^2 = 63$ $a \cdot 9 = 63 \implies a = \frac{63}{9} = 7$

Step 5: Find the 5th and 7th terms

The formula for the 5th term is: $T_5 = a \cdot r^4$ Substitute $a = 7$ and $r = 3$: $T_5 = 7 \cdot 3^4 = 7 \cdot 81 = 567 \; \text{(correct)}$

The formula for the 7th term is: $T_7 = a \cdot r^{7-1} = a \cdot r^6$ $T_7 = 7 \cdot 3^6 = 7 \cdot 729 = 5103$

Step 6: Sum of the 5th and 7th terms

$T_5 + T_7 = 567 + 5103 = 5670$

Final Answer:

The sum of the 5th and 7th terms is: $\boxed{5670}$

Do you want more details or help with related problems? Here are some related questions to explore:

1. How do you derive the sum of the first $n$ terms of a GP?
2. What is the relationship between the common ratio and the terms in a GP?
3. How do you solve for terms in a GP if $r$ is a fraction?
4. How do you solve a similar problem for an arithmetic progression (AP)?
5. How can the formula for $T_n$ be modified for complex numbers?

Tip: Always verify your solution by substituting back into the equations!