Let $\rm S_n$ be the sum of n terms of the arithmetic series. It is given that,

$S_6 = 183$ and $S_9 = 369$

We know, $S_n = \frac{n}{2} [2a + (n-1)d]$

Now,
$S_6 = \frac{6}{2} [2a + (6-1)d]$

$or, 183 =  3 × [2a + 5d]

$or, \frac{183}{3} = 2a + 5d$

$or, 61 = 2a + 5d$

$or, a = \frac{61 - 5d}{2}$ ----- (i)

And,
$S_9 = \frac{9}{2} [2a + (9-1)d]$

$or, 369 = \frac{9}{2} [2a + 8d]$

$or, \frac{369}{9} = \frac{1}{2} × 2[a +4d]$

$or, 41 = a + 4d$

$or, a = 41 - 4d$ ----- (ii)

From equations (i) and (ii), we get,
$or, \frac{61 - 5d}{2} = 41 - 4d$

$or, 61 - 5d = 2(41-4d)$

$or, 61 - 5d = 82 - 8d$

$or, 8d - 5d = 82 - 61$

$or, 3d = 21$

$\therefore d = 7$

Put the value of d in equation (ii), and we get,

$or, a = 41 - 4×7$

$or, a = 41 - 28$

$\therefore a = 13$

Hence, the required common difference and the first term of the arithmetic series are 7 and 13, respectively.