This is straightforward using Linq to Xml:
let recordTypeToXml (recordType: seq<ARecordType>) =
XElement(XName.Get "Root",
recordType |> Seq.map (fun {Aaa=a; Bbb=b; Ccc=c} ->
XElement(XName.Get "Row",
XAttribute(XName.Get "Aaa", a),
XAttribute(XName.Get "Bbb", b),
XAttribute(XName.Get "Ccc", c))))
You can shorten this a bit by wrapping some of the method calls with operators or functions. For example, with these two helper functions:
let Element name (content: seq<_>) = XElement(XName.Get name, content)
let Attr name value = XAttribute(XName.Get name, value)
it's much cleaner:
let recordTypeToXml (recordType: seq<ARecordType>) =
Element "Root"
[ for {Aaa=a; Bbb=b; Ccc=c} in recordType ->
Element "Row"
[ Attr "Aaa" a
Attr "Bbb" b
Attr "Ccc" c ] ]