Find the sum of 10 terms of the series : `96-48+24 . . . . . . . . . .` .

Answer

To find the sum of the first 10 terms of the series \(96, -48, 24, \ldots\), we can follow these steps:

### Step 1: Identify the first term and common ratio
The first term \(a\) of the series is:
\[ a = 96 \]

To find the common ratio \(r\), we can use the second term:
\[ r = \frac{a_2}{a_1} = \frac{-48}{96} = -\frac{1}{2} \]

### Step 2: Verify the common ratio
We can also verify the common ratio using the third term:
\[ r = \frac{a_3}{a_2} = \frac{24}{-48} = -\frac{1}{2} \]
This confirms that the series is a geometric progression (GP) with \(r = -\frac{1}{2}\).

### Step 3: Use the formula for the sum of the first \(n\) terms of a GP
The formula for the sum of the first \(n\) terms of a GP is:
\[ S_n = \frac{a(1 - r^n)}{1 - r} \]
where \(n\) is the number of terms.

### Step 4: Substitute the values into the formula
For our case, we need to find the sum of the first 10 terms (\(n = 10\)):
\[ S_{10} = \frac{96 \left(1 - \left(-\frac{1}{2}\right)^{10}\right)}{1 - \left(-\frac{1}{2}\right)} \]

### Step 5: Simplify the denominator
Calculating the denominator:
\[ 1 - \left(-\frac{1}{2}\right) = 1 + \frac{1}{2} = \frac{3}{2} \]

### Step 6: Calculate \(r^{10}\)
Now, calculate \(\left(-\frac{1}{2}\right)^{10}\):
\[ \left(-\frac{1}{2}\right)^{10} = \frac{1}{1024} \]

### Step 7: Substitute back into the sum formula
Now substitute back into the formula:
\[ S_{10} = \frac{96 \left(1 - \frac{1}{1024}\right)}{\frac{3}{2}} = \frac{96 \left(\frac{1024 - 1}{1024}\right)}{\frac{3}{2}} = \frac{96 \cdot \frac{1023}{1024}}{\frac{3}{2}} \]

### Step 8: Simplify the expression
This can be simplified as follows:
\[ S_{10} = 96 \cdot \frac{1023}{1024} \cdot \frac{2}{3} = \frac{192 \cdot 1023}{3072} \]

### Step 9: Final simplification
Now, simplifying:
\[ S_{10} = \frac{1023 \cdot 64}{16} = \frac{1023}{16} \]

Thus, the sum of the first 10 terms of the series is:
\[ S_{10} = \frac{1023}{16} \]