Counting up!

$Using \space the \space formula \space for \space sum \space of \space terms \space in \space an \space arithmetic \space sequence \space i.e.. S_{n} \space =\space \frac{n}{2}\space [2a \space + \space (n-1)d]$ $where \space S_{n}\space=\space sum \space of \space the \space terms\space, n \space= \space number \space of \space terms\space, \space a \space=\space first \space term \space, \space d \space=\space common \space difference \space$

$Here \space S_{n} \space = \space 120 \space, \space n \space=\space 6 \space,\space a\space=\space 10$

$\rightarrow \space 120\space=\space \frac{6}{2} \space[ 2\times 10 \space +\space 5d]$

$\rightarrow \space 120 \space=\space 3 \space (20\space+\space5d)$

$\rightarrow \space 5d \space = \space 20$

$\Rightarrow \space \boxed{\therefore \space d \space = \space 4}$

The mean of the 6 terms must be 20, since 20*6 = 120. Therefore, the last term is 30 (since the first and last term are equidistant from the mean). The final term will be 10+5x = 30. (30-10)/5 = 4.

$\begin{aligned} (a) + (a+d) + (a + 2d) + (a + 3d) + (a + 4d) + (a + 5d) &= 120 \\ (10) + (10 +d) + (10 + 2d) + (10 + 3d) + (10 + 4d) + (10 + 5d) &= 120 \quad [a = 10] \\ 60 + 15d &= 120 \\ d &= \boxed{4} \end{aligned}$