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2 answer?

Edit: According to Wikipedia, $A_4$ is not supersolvable, but $A_4 \cong (\mathbb{G}_a \rtimes \mathbb{G}_m)(\mathbb{F}_4)$. I think this answers your question in the negative.

Original response: What definition of unipotent algebraic group are you using? If you're only looking at groups over an algebraically closed field, then unipotent groups are defined in SGA3 Exp. 17 by the existence of a composition series with quotients given by subgroups of the additive group. The description of the central series is given in loc. cit. Theorem 3.5, and agrees with your conjecture.

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What definition of unipotent algebraic group are you using? If you're only looking at groups over an algebraically closed field, then unipotent groups are defined in SGA3 Exp. 17 by the existence of a composition series with quotients given by subgroups of the additive group. The description of the central series is given in loc. cit. Theorem 3.5, and agrees with your conjecture.