Arithmetic sequence

The terms $a_1,a_2,a_3$ form an arithmetic sequence whose sum is $18$. The terms $a_1+1,a_2,a_3+2$ in that order, form a geometric sequence. Find the sum of all possible values for $a_1$.
Source: NCTM Mathematics Teacher, February 2006

SOLUTION
Let $d$ represent the common difference of the arithmetic sequence
$a_1=a_1$
$a_2=a_1+d$
$a_3=a_1+2d$
$a_1+a_2+a_3=a_1+(a_1+d)+(a_1+2d)$
$=3a_1+3d$
$18=3a_1+3d$
$a_1+d=6\qquad\qquad (1)$
$a_2=6$
$a_3=a_1+2d$
$=a_1+d+d$
$=6+d$
In summary, the terms of the arithmetic sequence are
$a_1=a_1$
$a_2=6$
$a_3=6+d\qquad\qquad (2)$
The terms of the geometric sequence are
$a_1+1=a_1+1$
$a_2=6$
$a_3+2=6+d+2$
$=8+d$
By definition of geometric sequence
$\dfrac{6}{a_1+1}=\dfrac{8+d}{6}$
$36=(a_1+1)(8+d)$
The factors of $36$ are $1,2,3,4,6,9,12,18$, and $36$ from which we form the following sequences

Of these sequences only $3,6,12$ and $12,6,3$ are geometric sequences. Thus, all possible values for $a_1$ are $2$ and $11$. Their sum equals $2+11=13$.

Answer: $13$
Alternative solution
$a_1=a_1$
$a_2=6$
$a_3=6+d$ from Eq. $(2)$
$a_1+a_3=a_1+6+d$
$=(a_1+d)+6$
$=6+6$ from Eq. $(1)$
$=12\qquad\qquad\qquad (3)$
By definition of geometric sequence
$\dfrac{6}{a_1+1}=\dfrac{a_3+2}{6}$
Substitute the value of $a_3$ from Eq. $(3)$
$\dfrac{6}{a_1+1}=\dfrac{12-a_1+2}{6}$
$36=(a_1+1)(14-a_1)$
$=14a_1-a_1^2+14-a_1$
$a_1^2-13a_1+22=0$
$(a_1-2)(a_1-11)=0$
$a_1=2$ or $a_1=11$