The sum of first three terms of a G.P. is 15 and sum of next three terms is 120. Find the sum of first n terms.

Answer

To solve the problem step by step, we will follow the given information about the geometric progression (G.P.) and derive the necessary formulas.

### Step 1: Define the terms of the G.P.
Let the first term of the G.P. be \( a \) and the common ratio be \( r \). The first three terms of the G.P. can be expressed as:
- First term: \( a \)
- Second term: \( ar \)
- Third term: \( ar^2 \)

### Step 2: Set up the equation for the sum of the first three terms
According to the problem, the sum of the first three terms is given as 15:
\[ a + ar + ar^2 = 15 \]
Factoring out \( a \):
\[ a(1 + r + r^2) = 15 \quad \text{(Equation 1)} \]

### Step 3: Set up the equation for the sum of the next three terms
The next three terms of the G.P. are:
- Fourth term: \( ar^3 \)
- Fifth term: \( ar^4 \)
- Sixth term: \( ar^5 \)

The sum of these terms is given as 120:
\[ ar^3 + ar^4 + ar^5 = 120 \]
Factoring out \( ar^3 \):
\[ ar^3(1 + r + r^2) = 120 \quad \text{(Equation 2)} \]

### Step 4: Relate the two equations
From Equation 1, we have:
\[ 1 + r + r^2 = \frac{15}{a} \]
Substituting this into Equation 2 gives:
\[ ar^3 \left(\frac{15}{a}\right) = 120 \]
This simplifies to:
\[ 15r^3 = 120 \]
Dividing both sides by 15:
\[ r^3 = 8 \]
Taking the cube root:
\[ r = 2 \]

### Step 5: Substitute \( r \) back to find \( a \)
Now substitute \( r = 2 \) back into Equation 1:
\[ a(1 + 2 + 2^2) = 15 \]
Calculating \( 1 + 2 + 4 = 7 \):
\[ a \cdot 7 = 15 \]
Thus:
\[ a = \frac{15}{7} \]

### Step 6: Find the sum of the first \( n \) terms of the G.P.
The formula for the sum of the first \( n \) terms of a G.P. is given by:
\[ S_n = a \frac{r^n - 1}{r - 1} \]
Substituting \( a = \frac{15}{7} \) and \( r = 2 \):
\[ S_n = \frac{15}{7} \cdot \frac{2^n - 1}{2 - 1} \]
This simplifies to:
\[ S_n = \frac{15}{7} (2^n - 1) \]

### Final Answer
The sum of the first \( n \) terms of the G.P. is:
\[ S_n = \frac{15}{7} (2^n - 1) \]