Fourth in a sequence...

[The Highlander]

Here's a quite interesting (I thought) little puzzle...

What should be on the fourth card?

​

Be sure to post your explanation (in a spoiler?) of how you arrived at what you propose the fourth card should contain.
(There is a specific sequence that determines its content.)

Please don't reply with the answer if you saw the program that the problem was broadcast on. ​

I couldn't get it in the allotted time (40 seconds) and, even after pausing the recording for a few minutes, I was still stumped! lol. (I was a bit tickled when I eventually saw the answer; hence the post here. )

Anyone got any idea what the solution is?

 

[logistic_guy]

Here's a quite interesting (I thought) little puzzle...

What should be on the fourth card?

View attachment 39052​

Be sure to post your explanation (in a spoiler?) of how you arrived at what you propose the fourth card should contain.
(There is a specific sequence that determines its content.)

Please don't reply with the answer if you saw the program that the problem was broadcast on. ​

I couldn't get it in the allotted time (40 seconds) and, even after pausing the recording for a few minutes, I was still stumped! lol. (I was a bit tickled when I eventually saw the answer; hence the post here. )

Anyone got any idea what the solution is?

Spoiler

If no calculations are involved, I think that the pattern:
\(\displaystyle 1\times 51 \)
\(\displaystyle 3\times 18 \)
\(\displaystyle 3\times 19 \)

If calculations are involved, I think that it would make sense to choose this pattern:

\(\displaystyle 1\times 52 \)
\(\displaystyle 3\times 18 \)
\(\displaystyle 3\times 19 \)

Why?

We start with \(\displaystyle 3\times 12 = 36\)
then
\(\displaystyle 2\times 19 = 38 \)

It looks like increase \(\displaystyle 2\) one time, then increase \(\displaystyle 1\) two times, then again \(\displaystyle 2\) one time, then \(\displaystyle 3\) two times, then again \(\displaystyle 2\) one time and the patterns continue like this.

\(\displaystyle 36 \ \ 40 \ \ 48 \ \ 52\)
\(\displaystyle 38 \ \ 42 \ \ 50 \ \ 54\)
\(\displaystyle 39 \ \ 45 \ \ 51 \ \ 57\)

where I chose \(\displaystyle 52 = 1\times 52\)