Find the sum of last 8 terms of the A.P. -12,-10,-8,……..,58.

Options

A. `400`
B. `428`
C. `308`
D. `408`

Answer

To find the sum of the last 8 terms of the arithmetic progression (A.P.) given by -12, -10, -8, ..., 58, we can follow these steps:

### Step 1: Identify the first term and common difference
The first term \( a \) of the A.P. is -12, and the common difference \( d \) can be calculated as follows:
\[ d = -10 - (-12) = -10 + 12 = 2 \]

### Step 2: Determine the last term
The last term \( l \) of the A.P. is given as 58.

### Step 3: Find the number of terms in the A.P.
We can use the formula for the nth term of an A.P.:
\[ l = a + (n-1) \cdot d \]
Substituting the known values:
\[ 58 = -12 + (n-1) \cdot 2 \]
Now, solve for \( n \):
\[ 58 + 12 = (n-1) \cdot 2 \]
\[ 70 = (n-1) \cdot 2 \]
\[ n-1 = 35 \]
\[ n = 36 \]

### Step 4: Find the sum of the last 8 terms
The last 8 terms of the A.P. can be considered as the first 8 terms of the A.P. starting from the last term 58 and moving backwards. The first term of this new A.P. is 58, and the common difference is -2 (since we are moving backwards).

### Step 5: Use the formula for the sum of the first \( n \) terms
The formula for the sum of the first \( n \) terms of an A.P. is:
\[ S_n = \frac{n}{2} \cdot (2a + (n-1)d) \]
Here, \( n = 8 \), \( a = 58 \), and \( d = -2 \).

Substituting the values:
\[ S_8 = \frac{8}{2} \cdot (2 \cdot 58 + (8-1)(-2)) \]
\[ S_8 = 4 \cdot (116 - 14) \]
\[ S_8 = 4 \cdot 102 \]
\[ S_8 = 408 \]

### Conclusion
The sum of the last 8 terms of the A.P. is **408**.

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