The sum of 30 terms of a series in A.P., whose last term is 98, is 1635. Find the first term and the common difference.

Answer

To solve the problem step by step, we will use the formulas related to arithmetic progressions (A.P.).

### Step 1: Identify the given information
We are given:
- The number of terms, \( n = 30 \)
- The last term, \( l = 98 \)
- The sum of the series, \( S_n = 1635 \)

### Step 2: Use the formula for the last term of an A.P.
The formula for the \( n \)-th term (or last term in this case) of an A.P. is given by:
\[ l = a + (n - 1)d \]
Substituting the known values:
\[ 98 = a + (30 - 1)d \]
This simplifies to:
\[ 98 = a + 29d \quad \text{(Equation 1)} \]

### Step 3: Use the formula for the sum of the first \( n \) terms of an A.P.
The formula for the sum of the first \( n \) terms is:
\[ S_n = \frac{n}{2} \times (a + l) \]
Substituting the known values:
\[ 1635 = \frac{30}{2} \times (a + 98) \]
This simplifies to:
\[ 1635 = 15(a + 98) \]
Dividing both sides by 15:
\[ 109 = a + 98 \]
Rearranging gives us:
\[ a = 109 - 98 \]
Thus:
\[ a = 11 \quad \text{(Equation 2)} \]

### Step 4: Substitute the value of \( a \) back into Equation 1
Now, we substitute \( a = 11 \) into Equation 1:
\[ 98 = 11 + 29d \]
Rearranging gives:
\[ 98 - 11 = 29d \]
So:
\[ 87 = 29d \]
Dividing both sides by 29 gives:
\[ d = \frac{87}{29} = 3 \]

### Final Answer
Thus, the first term \( a \) is \( 11 \) and the common difference \( d \) is \( 3 \).

### Summary of Results
- First term \( a = 11 \)
- Common difference \( d = 3 \)