How many terms of G.P. `3,3^(2),3^(3),……` are needed to give the sum 120?

Explanation

To find how many terms of the geometric progression (G.P.) \(3, 3^2, 3^3, \ldots\) are needed to give the sum of 120, we can use the formula for the sum of the first \(n\) terms of a G.P.

### Step-by-step Solution:

1. **Identify the first term (A) and the common ratio (r):**
- The first term \(A = 3\).
- The common ratio \(r = \frac{3^2}{3} = 3\).

2. **Use the formula for the sum of the first \(n\) terms of a G.P.:**
The formula for the sum \(S_n\) of the first \(n\) terms of a G.P. is given by:
\[ S_n = \frac{A(r^n - 1)}{r - 1} \]
Substituting the values we have:
\[ S_n = \frac{3(3^n - 1)}{3 - 1} = \frac{3(3^n - 1)}{2} \]

3. **Set the sum equal to 120:**
We need to find \(n\) such that:
\[ \frac{3(3^n - 1)}{2} = 120 \]

4. **Multiply both sides by 2 to eliminate the fraction:**
\[ 3(3^n - 1) = 240 \]

5. **Divide both sides by 3:**
\[ 3^n - 1 = 80 \]

6. **Add 1 to both sides:**
\[ 3^n = 81 \]

7. **Recognize that \(81\) is a power of \(3\):**
\[ 3^n = 3^4 \]

8. **Equate the exponents:**
Since the bases are the same, we can set the exponents equal to each other:
\[ n = 4 \]

### Final Answer:
Thus, **4 terms** of the G.P. are needed to give the sum of 120.

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