Find the sums given below:
`34 + 32 + 30 + ...+ 10`

Answer

To find the sum of the arithmetic progression given by the series \(34 + 32 + 30 + \ldots + 10\), we will follow these steps:

### Step 1: Identify the first term and the common difference
- The first term \(a\) is \(34\).
- The common difference \(d\) can be calculated as follows:
\[ d = 32 - 34 = -2 \]

### Step 2: Identify the last term
- The last term \(l\) is given as \(10\).

### Step 3: Use the formula for the \(n\)-th term of an arithmetic progression
The formula for the \(n\)-th term \(T_n\) of an arithmetic progression is:
\[ T_n = a + (n - 1) \cdot d \]
We know \(T_n = 10\), \(a = 34\), and \(d = -2\). Substituting these values into the formula gives:
\[ 10 = 34 + (n - 1)(-2) \]

### Step 4: Solve for \(n\)
Rearranging the equation:
\[ 10 = 34 - 2(n - 1) \]
\[ 10 = 34 - 2n + 2 \]
\[ 10 = 36 - 2n \]
Subtracting \(36\) from both sides:
\[ 10 - 36 = -2n \]
\[ -26 = -2n \]
Dividing both sides by \(-2\):
\[ n = \frac{26}{2} = 13 \]

### Step 5: Use the formula for the sum of the first \(n\) terms
The formula for the sum \(S_n\) of the first \(n\) terms of an arithmetic progression is:
\[ S_n = \frac{n}{2} \cdot (a + l) \]
Substituting \(n = 13\), \(a = 34\), and \(l = 10\):
\[ S_{13} = \frac{13}{2} \cdot (34 + 10) \]
Calculating \(34 + 10\):
\[ 34 + 10 = 44 \]
Now substituting back into the sum formula:
\[ S_{13} = \frac{13}{2} \cdot 44 \]
Calculating \(\frac{13 \cdot 44}{2}\):
\[ S_{13} = \frac{572}{2} = 286 \]

### Conclusion
The sum of the series \(34 + 32 + 30 + \ldots + 10\) is \(286\).

---