Question

The sum of 11 terms of an Appointment is 891 find the 1st 28th and 45th terms if the common difference is 15

Answer 1

To find the first, 28th, and 45th terms of an arithmetic sequence (AP), we first need to know the formula for the \( n \)-th term of an AP and the sum of the first \( n \) terms.

1. **Formula for the \( n \)-th term**:
\[
a_n = a + (n - 1)d
\]
where:
- \( a \) is the first term,
- \( d \) is the common difference,
- \( n \) is the term number.

2. **Formula for the sum of the first \( n \) terms**:
\[
S_n = \frac{n}{2} \times (2a + (n - 1)d)
\]
where \( S_n \) is the sum of the first \( n \) terms.

In this problem, we know:
- The sum of the first 11 terms \( S_{11} = 891 \).
- The common difference \( d = 15 \).
- We are trying to find \( a \), the first term.

### Step 1: Using the sum formula
Substituting the values into the sum formula for the first 11 terms:

\[
S_{11} = \frac{11}{2} \times (2a + (11 - 1) \times 15)
\]

This simplifies to:

\[
891 = \frac{11}{2} \times (2a + 10 \times 15)
\]

Calculating \( 10 \times 15 \):

\[
10 \times 15 = 150
\]

So the equation becomes:

\[
891 = \frac{11}{2} \times (2a + 150)
\]

Multiplying both sides by 2 to eliminate the fraction:

\[
1782 = 11 \times (2a + 150)
\]

Now, dividing both sides by 11:

\[
162 = 2a + 150
\]

Subtracting 150 from both sides:

\[
12 = 2a
\]

Dividing by 2 gives:

\[
a = 6
\]

### Step 2: Finding the first, 28th, and 45th terms
Now that we have \( a = 6 \) and \( d = 15 \), we can find the terms.

1. **First term ( \( a_1 \) )**:
\[
a_1 = a = 6
\]

2. **28th term ( \( a_{28} \) )**:
\[
a_{28} = a + (28 - 1)d = 6 + (27 \times 15)
\]
\[
= 6 + 405 = 411
\]

3. **45th term ( \( a_{45} \) )**:
\[
a_{45} = a + (45 - 1)d = 6 + (44 \times 15)
\]
\[
= 6 + 660 = 666
\]

### Summary of Results
- First term \( a_1 = 6 \)
- 28th term \( a_{28} = 411 \)
- 45th term \( a_{45} = 666 \)