Find the sum of the series 81 -27 +9 - ...... `-(1)/(27)` .

Answer

To find the sum of the series \( 81 - 27 + 9 - \ldots - \frac{1}{27} \), we can identify that this is a geometric progression (GP). Let's break down the solution step by step.

### Step 1: Identify the first term and the common ratio
The first term \( a \) of the series is:
\[ a = 81 \]
The common ratio \( r \) can be found by dividing the second term by the first term:
\[ r = \frac{-27}{81} = -\frac{1}{3} \]
We can also verify this by dividing the third term by the second term:
\[ r = \frac{9}{-27} = -\frac{1}{3} \]

### Step 2: Find the number of terms in the series
The last term of the series is given as \( -\frac{1}{27} \). We can use the formula for the \( n \)-th term of a geometric progression:
\[ a_n = a \cdot r^{n-1} \]
Setting this equal to the last term:
\[ -\frac{1}{27} = 81 \cdot \left(-\frac{1}{3}\right)^{n-1} \]
To simplify, we can express \( -\frac{1}{27} \) as:
\[ -\frac{1}{27} = -\frac{1}{3^3} \]
Thus, we have:
\[ -\frac{1}{3^3} = 81 \cdot \left(-\frac{1}{3}\right)^{n-1} \]
Now, we can express \( 81 \) as \( 3^4 \):
\[ -\frac{1}{3^3} = 3^4 \cdot \left(-\frac{1}{3}\right)^{n-1} \]
This simplifies to:
\[ -\frac{1}{3^3} = -3^{4 - (n-1)} \]
This leads to:
\[ 3^3 = 3^{5-n} \]
Equating the exponents gives:
\[ 3 = 5 - n \implies n = 2 \]

### Step 3: Use the formula for the sum of the first \( n \) terms of a GP
The formula for the sum \( S_n \) of the first \( n \) terms of a geometric series is:
\[ S_n = a \cdot \frac{1 - r^n}{1 - r} \]
Substituting the values we have:
\[ S_n = 81 \cdot \frac{1 - \left(-\frac{1}{3}\right)^n}{1 - \left(-\frac{1}{3}\right)} \]
Now substituting \( n = 8 \):
\[ S_8 = 81 \cdot \frac{1 - \left(-\frac{1}{3}\right)^8}{1 + \frac{1}{3}} = 81 \cdot \frac{1 - \frac{1}{6561}}{\frac{4}{3}} \]
This simplifies to:
\[ S_8 = 81 \cdot \frac{3}{4} \cdot \left(1 - \frac{1}{6561}\right) \]

### Step 4: Calculate the sum
Calculating \( 1 - \frac{1}{6561} \):
\[ 1 - \frac{1}{6561} = \frac{6560}{6561} \]
Thus,
\[ S_8 = 81 \cdot \frac{3}{4} \cdot \frac{6560}{6561} \]
Calculating \( 81 \cdot \frac{3}{4} = 60.75 \):
\[ S_8 = 60.75 \cdot \frac{6560}{6561} \]
This is approximately \( 60.74 \).

### Final Answer
The sum of the series \( 81 - 27 + 9 - \ldots - \frac{1}{27} \) is approximately \( 60.74 \).