Find the dissimilar term in the given sequence: ACA, BFB, CJC, DLD, EOE

Let's analyze the pattern of each three-letter term:
The first and third letters are identical and follow the sequence of consecutive English alphabets: $A,B,C,D,E$.
The middle letter is determined by adding an increasing number to the position of the outer letters:

* ACA: $A=1$, Middle letter $=1+2=3\to C$
* BFB: $B=2$, Middle letter $=2+4=6\to F$
* CJC: $C=3$, Middle letter $=3+7=10\to J$
* DLD: $D=4$, Middle letter $=4+8=12\to L$
* EOE: $E=5$, Middle letter $=5+10=15\to O$

Let's observe the differences added to the first letter to get the middle letter: $+2,+4,+7,+8,+10$. This sequence does not have a consistent progression.
Alternatively, look at the position of the middle letters directly:

* C $=3$
* F $=6$
* J $=10$
* L $=12$
* O $=15$

The gaps between the middle letters are:

* From C to F: $6-3=+3$
* From F to J: $10-6=+4$
* From J to L: $12-10=+2$
* From L to O: $15-12=+3$

If the middle letters followed a uniform arithmetic progression with a common difference of $+3$ ($3,6,9,12,15$), the middle letters would be C, F, I, L, O.
The third term would be CIC instead of CJC. Thus, CJC is the dissimilar term.