Correct Answer: 3

Solution : Given: The sum of 10 terms of the arithmetic series is 390.
So, $n=10$
The third term of the series is 19.
Let the first term of the series be $a$.
We know, sum of arithmetic progression (A.P.) = $\frac{n}{2}[2a+(n-1)d]$
$n^{th}$ term = $a+(n-1)d$
So, $3^{rd}$ term ⇒ $a+(3-1)d=19$
⇒ $a+2d=19$ ----------------------------(1)
Sum of 10 terms = $\frac{10}{2}[2a+(10-1)d] =390$
⇒ $2a+9d=78$ ---------(2)
Multiplying 9 with equation (1) and 2 with equation (2), we get,
⇒ $9a+18d=171$ ------------------------(3)
⇒ $4a+18d=156$ ------------------------(4)
Subtracting (4) from (3), we get,
$5a=15$
$\therefore a=3$
Hence, the correct answer is 3.