The problem reads:

The 3rd and 8th terms of a G.P. are 4 and 128 respectively. Find the G.P.

Solution:

For a geometric progression (G.P.), the $n$-th term is given by: $a_n = a \cdot r^{n-1}$ where $a$ is the first term, and $r$ is the common ratio.

Step 1: Express the given terms using the formula

• For the 3rd term ($a_3 = 4$): a \cdot r^{3-1} = a \cdot r^2 = 4 \tag{1}

• For the 8th term ($a_8 = 128$): a \cdot r^{8-1} = a \cdot r^7 = 128 \tag{2}

Step 2: Divide equations to eliminate $a$

Dividing $(2)$ by $(1)$: $\frac{a \cdot r^7}{a \cdot r^2} = \frac{128}{4}$ $r^{7-2} = 32 \implies r^5 = 32$

Step 3: Solve for $r$

Take the fifth root of both sides: $r = \sqrt[5]{32} = 2$

Step 4: Solve for $a$

Substitute $r = 2$ into equation $(1)$: $a \cdot (2)^2 = 4$ $a \cdot 4 = 4 \implies a = 1$

Step 5: Write the G.P.

The G.P. has the first term $a = 1$ and common ratio $r = 2$. Therefore, the G.P. is: $1, 2, 4, 8, 16, 32, 64, 128, \dots$

Final Answer:

The geometric progression is: $1, 2, 4, 8, 16, 32, 64, 128, \dots$

Would you like more explanation or examples?

Related Questions:

1. What is the formula for the sum of $n$ terms of a G.P.?
2. How does the common ratio $r$ affect the progression of terms in a G.P.?
3. Can a G.P. have a common ratio that is a fraction? What happens in that case?
4. How would the solution change if the terms were $a_3 = 4$ and $a_8 = -128$?
5. What is the significance of the first term in defining a G.P.?

Tip:

When solving problems involving geometric progressions, dividing the equations for different terms often simplifies the calculation by eliminating the first term $a$.