If seven times the seventh term of an A.P is equal to eleven times its eleventh term , show that its eighteenth term is zero .

Answer

To solve the problem, we need to show that the eighteenth term of an arithmetic progression (A.P.) is zero, given that seven times the seventh term is equal to eleven times the eleventh term.

### Step-by-Step Solution:

1. **Understanding the Terms of A.P.:**
The nth term of an A.P. can be expressed as:
\[ A_n = A + (n - 1)d \]
where \( A \) is the first term, \( d \) is the common difference, and \( n \) is the term number.

2. **Expressing the 7th and 11th Terms:**
- The 7th term \( A_7 \) is:
\[ A_7 = A + (7 - 1)d = A + 6d \]
- The 11th term \( A_{11} \) is:
\[ A_{11} = A + (11 - 1)d = A + 10d \]

3. **Setting Up the Given Equation:**
According to the problem, we have:
\[ 7 \times A_7 = 11 \times A_{11} \]
Substituting the expressions for \( A_7 \) and \( A_{11} \):
\[ 7(A + 6d) = 11(A + 10d) \]

4. **Expanding Both Sides:**
Expanding both sides gives:
\[ 7A + 42d = 11A + 110d \]

5. **Rearranging the Equation:**
Rearranging the equation to bring all terms involving \( A \) and \( d \) to one side:
\[ 7A + 42d - 11A - 110d = 0 \]
Simplifying this, we get:
\[ -4A - 68d = 0 \]

6. **Factoring Out Common Terms:**
Factoring out \(-4\) from the equation:
\[ 4A + 68d = 0 \]

7. **Dividing by 4:**
Dividing the entire equation by 4:
\[ A + 17d = 0 \]

8. **Finding the Eighteenth Term:**
The eighteenth term \( A_{18} \) can be expressed as:
\[ A_{18} = A + (18 - 1)d = A + 17d \]
Substituting \( A + 17d = 0 \) into this equation:
\[ A_{18} = 0 \]

Thus, we have shown that the eighteenth term of the A.P. is zero.