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

Let's denote the first term of the arithmetic progression (AP) as $a$ and the common difference as $d$.

The $n$-th term of an AP is given by: $a_n = a + (n-1)d$

Given that 7 times the 7th term is equal to 11 times the 11th term, we can write: $7 \times a_7 = 11 \times a_{11}$

Substitute the expressions for the 7th and 11th terms: $7 \times (a + 6d) = 11 \times (a + 10d)$

Expand and simplify: $7a + 42d = 11a + 110d$

Rearrange to isolate terms involving $a$ and $d$: $7a - 11a = 110d - 42d$ $-4a = 68d$ $a = -17d$

Now, we need to find the 18th term of the AP: $a_{18} = a + 17d$

Substitute $a = -17d$: $a_{18} = -17d + 17d$ $a_{18} = 0$

Therefore, the 18th term of the AP is $0$.