Find the `11^(t h)`from the last term (towards the first term) of the AP : `10 ,\ 7,\ 4,\ dot\ dot\ dot\ ,\ \ -62.`

Answer

To find the 11th term from the last of the arithmetic progression (AP) given by the sequence: \(10, 7, 4, \ldots, -62\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the AP and its terms**:
The given AP is \(10, 7, 4, \ldots, -62\). The first term \(a = 10\) and the last term is \(-62\).

2. **Calculate the common difference \(d\)**:
The common difference \(d\) can be calculated as:
\[ d = 7 - 10 = -3 \]
So, \(d = -3\).

3. **Determine the number of terms \(n\) in the AP**:
The formula for the \(n\)th term of an AP is given by:
\[ a_n = a + (n-1)d \]
Setting \(a_n = -62\), we can solve for \(n\):
\[ -62 = 10 + (n-1)(-3) \]
Rearranging gives:
\[ -62 - 10 = (n-1)(-3) \\ -72 = (n-1)(-3) \\ n-1 = \frac{-72}{-3} \\ n-1 = 24 \\ n = 25 \]
Thus, there are \(25\) terms in the AP.

4. **Find the 11th term from the last**:
The 11th term from the last can be calculated as:
\[ \text{Position from the start} = n - 11 + 1 = 25 - 11 + 1 = 15 \]
So, we need to find the 15th term of the AP.

5. **Calculate the 15th term**:
Using the formula for the \(n\)th term:
\[ a_{15} = a + (15-1)d \\ a_{15} = 10 + (14)(-3) \\ a_{15} = 10 - 42 \\ a_{15} = -32 \]

### Final Answer:
The 11th term from the last of the AP is \(-32\).

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