Answer:
The general form of an arithmetic progression (A.P.) is given by \( a_n = a_1 + (n-1)d \), where:
- \( a_n \) is the nth term,
- \( a_1 \) is the first term,
- \( n \) is the number of terms,
- \( d \) is the common difference.
Given:
- \( a_1 = 3 \) (the first term),
- \( a_{50} = 9 \) (the 50th term),
- \( a_{\text{last}} = 81 \) (the last term).
For an A.P., \( a_n = a_1 + (n-1)d \), so we can set up equations using the given information:
1. For the 50th term:
\[ a_{50} = a_1 + (50-1)d = 3 + 49d = 9 \]
2. For the last term:
\[ a_{\text{last}} = a_1 + (n-1)d = 3 + (n-1)d = 81 \]
Now, solve these equations to find \( d \) and \( n \).
From the first equation:
\[ 3 + 49d = 9 \]
\[ 49d = 6 \]
\[ d = \frac{6}{49} \]
Now substitute \( d \) into the second equation:
\[ 3 + (n-1)\left(\frac{6}{49}\right) = 81 \]
Solve for \( n \):
\[ (n-1)\left(\frac{6}{49}\right) = 78 \]
\[ n - 1 = \frac{78 \times 49}{6} \]
\[ n - 1 = 637 \]
\[ n = 638 \]
Therefore, the number of terms in the progression is 638 when the last term is 81.