For given function type `Fn`, and any type `A` (any in this context means we don't restrict the type, and I don't have in mind any type 😉) create a generic type which will take `Fn` as the first argument, `A` as the second, and will produce function type `G` which will be the same as `Fn` but with appended argument `A` as a last one.
For example,
```typescript
type Fn = (a: number, b: string) => number
type Result = AppendArgument
// expected be (a: number, b: string, x: boolean) => number
```
> This question is ported from the [original article](https://dev.to/macsikora/advanced-typescript-exercises-question-4-495c) by [@maciejsikora](https://github.com/maciejsikora)
