Each term of A.P is doubled. Is the resulting sequence also an A.P ? If it is write its first term.

Let the first term be $a$ and the common difference be $d$. The $n^{th}$ term of an A.P. is given by $a_n = a + (n-1)d$.
The $7^{th}$ term is $a_7 = a + (7-1)d = a + 6d$.
The $11^{th}$ term is $a_{11} = a + (11-1)d = a + 10d$.

The problem states that 7 times the $7^{th}$ term is equal to 11 times the $11^{th}$ term:

Substitute the expressions from Step 1:

Subtract $7a$ from both sides:

Subtract $110d$ from both sides:

Divide by 4:

The $18^{th}$ term is $a_{18} = a + (18-1)d = a + 17d$.

Thus, the $18^{th}$ term of the A.P. is $\boxed{0}$.