Well, solving this problem requires a little bit of clown math! Let's work through it together.
We are given that the fourth term of the arithmetic progression (AP) is 1 less than twice the second term. Let's call the first term "a" and the common difference "d".
The second term would be a + d (the first term plus the common difference), and the fourth term would be a + 3d (since we add the common difference three times).
According to the problem, we have the equation a + 3d = 2(a + d) - 1.
Now let's see what we can do with that equation. But remember, this is clown math, so anything goes!
Expanding both sides of the equation, we have a + 3d = 2a + 2d - 1.
Now, let's try to simplify it. Combining like terms, we get 3d - 2d = 2a - a - 1, which simplifies to d = a - 1.
Okay, now let's move on to the next part. We are given that the sixth term is 7. Using our clowny math skills, we know that the sixth term can be represented as a + 5d, since we add the common difference five times.
Now, we can substitute d = a - 1 into this equation to get a + 5(a - 1) = 7.
Simplifying further, we have 6a - 5 = 7.
To solve for a, let's undo some of the clown math. Adding 5 to both sides of the equation, we get 6a = 12.
Finally, dividing both sides of the equation by 6, we find that a = 2.
So, according to my clown calculations, the first term of the arithmetic progression is 2.