If 7 times the 7th term of an AP is equal to 11 times its 11th term, then its 18th term will be

Options

A. 7
B. 11
C. 18
D. 0

Explanation

To solve the problem step by step, we will use the properties of arithmetic progressions (AP).

### Step 1: Understand the terms of the AP
The nth term of an AP 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.

### Step 2: Write the expressions for the 7th and 11th terms
Using the formula for the nth term:
- The 7th term \( a_7 \) is:
\[ a_7 = a + (7 - 1)d = a + 6d \]
- The 11th term \( a_{11} \) is:
\[ a_{11} = a + (11 - 1)d = a + 10d \]

### Step 3: Set up the equation from the problem statement
According to the problem, 7 times the 7th term is equal to 11 times the 11th term:
\[ 7 \cdot a_7 = 11 \cdot a_{11} \]
Substituting the expressions we found:
\[ 7(a + 6d) = 11(a + 10d) \]

### Step 4: Expand both sides of the equation
Expanding both sides gives:
\[ 7a + 42d = 11a + 110d \]

### Step 5: Rearrange the equation
Rearranging the equation to isolate terms involving \( a \) and \( d \):
\[ 7a + 42d - 11a - 110d = 0 \]
This simplifies to:
\[ -4a - 68d = 0 \]

### Step 6: Factor out common terms
Factoring out -4 gives:
\[ 4a + 68d = 0 \]
Dividing through by 4:
\[ a + 17d = 0 \]
This implies:
\[ a = -17d \]

### Step 7: Find the 18th term
Now, we need to find the 18th term \( a_{18} \):
\[ a_{18} = a + (18 - 1)d = a + 17d \]
Substituting \( a = -17d \):
\[ a_{18} = -17d + 17d = 0 \]

### Conclusion
Thus, the 18th term of the AP is:
\[ \boxed{0} \]