The first and fifth terms of an A.P. are -14 and 2 respectively and the sum of its $n$ terms is 40 . The value of $n$ is

Let the first term of the arithmetic progression be $a$ and the common difference be $d$.

Given $a = -14$ and the fifth term $a_5 = a + 4d = 2$.

Substituting $a = -14$ into the equation $a + 4d = 2$:

$-14 + 4d = 2 \Rightarrow 4d = 16 \Rightarrow d = 4$.

The sum of $n$ terms of an A.P. is given by $S_n = \dfrac{n}{2} [2a + (n - 1)d]$.

Given $S_n = 40$, we have $40 = \dfrac{n}{2} [2(-14) + (n - 1)4]$.

$80 = n [-28 + 4n - 4]$.

$80 = n [4n - 32]$.

$80 = 4n^2 - 32n$.

$n^2 - 8n - 20 = 0$.

Factoring the quadratic equation: $(n - 10)(n + 2) = 0$.

Since $n$ must be a positive integer, $n = 10$.

Answer: 10