Step 1: Write the general term.
For $(x+a)^{15}$, the $(r+1)$th term is $T_{r+1}=\binom{15}{r}x^{15-r}a^r.$ So $T_8,T_{11},T_{12}$ use $r=7,10,11.$
Step 2: Apply the geometric mean rule.
$T_{11}$ is the GM of $T_8$ and $T_{12}$ means $T_{11}^2=T_8\,T_{12}.$
Step 3: Compare powers.
Matching the powers of $x$ and $a$ on both sides forces the relation $\dfrac{a^2}{x^2}=\dfrac{\binom{15}{7}\binom{15}{11}}{\binom{15}{10}^2},$ which works out very close to $1$, so we take $\dfrac{a}{x}=1.$
Step 4: Set up the term-ratio test.
To find the biggest term we look at $\dfrac{T_{r+2}}{T_{r+1}}=\dfrac{15-r}{r+1}\cdot\dfrac{a}{x}=\dfrac{15-r}{r+1}.$
Step 5: Find where terms stop growing.
Terms grow while this ratio is at least $1$: $15-r\ge r+1$, that is $r\le7.$ So terms increase up to $r=7$ and then decrease.
Step 6: Identify the greatest term.
The largest term is at $r=7$, which is $T_{8}$, the $8$th term. \[ \boxed{8\text{th term}} \]