Find the sum of G.P. :
`1-(1)/(3)+(1)/(3^(2))-(1)/(3^(3))+ . . . .. . . . .` to n terms.

Answer

To find the sum of the geometric progression (G.P.) given by the series:

\[ S_n = 1 - \frac{1}{3} + \frac{1}{3^2} - \frac{1}{3^3} + \ldots \text{ (to n terms)} \]

we will use the formula for the sum of the first n terms of a G.P., which is:

\[ S_n = A \frac{1 - r^n}{1 - r} \]

where:
- \( A \) is the first term of the G.P.
- \( r \) is the common ratio.

### Step 1: Identify the first term (A) and the common ratio (r)

The first term \( A \) is:

\[ A = 1 \]

The common ratio \( r \) can be found by dividing the second term by the first term:

\[ r = \frac{-\frac{1}{3}}{1} = -\frac{1}{3} \]

### Step 2: Substitute A and r into the sum formula

Now we can substitute \( A \) and \( r \) into the sum formula:

\[ S_n = 1 \cdot \frac{1 - \left(-\frac{1}{3}\right)^n}{1 - \left(-\frac{1}{3}\right)} \]

### Step 3: Simplify the denominator

The denominator simplifies as follows:

\[ 1 - \left(-\frac{1}{3}\right) = 1 + \frac{1}{3} = \frac{3}{3} + \frac{1}{3} = \frac{4}{3} \]

### Step 4: Substitute the simplified denominator back into the formula

Now substituting back into the formula gives:

\[ S_n = \frac{1 - \left(-\frac{1}{3}\right)^n}{\frac{4}{3}} \]

### Step 5: Simplify the expression

To simplify, we multiply by the reciprocal of the denominator:

\[ S_n = \left(1 - \left(-\frac{1}{3}\right)^n\right) \cdot \frac{3}{4} \]

Thus, we have:

\[ S_n = \frac{3}{4} \left(1 - \left(-\frac{1}{3}\right)^n\right) \]

### Final Result

The sum of the G.P. to n terms is:

\[ S_n = \frac{3}{4} \left(1 - \left(-\frac{1}{3}\right)^n\right) \]

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