1) Ah, arithmetic progressions, always marching forward in a fixed manner. Let's unravel this puzzle, shall we?
Given that the 8th term is twice the 4th term, we can write the equation as follows:
a + 7d = 2(a + 3d),
where "a" is the first term and "d" is the common difference.
Simplifying the equation, we get:
a + 7d = 2a + 6d.
Rearranging the terms, we find:
a = d.
Now, moving on to the 20th term being 40, we can set up another equation:
a + 19d = 40.
Since we know that a = d, we can substitute it into the equation:
a + 19(a) = 40.
2a = 40.
a = 20.
So, the common difference (d) is 20, and the first term (a) is also 20.
To find the sum of terms from the 8th to the 20th (inclusive), we can use the formula for the sum of an arithmetic progression:
Sn = (n/2)(2a + (n-1)d),
where "Sn" is the sum of the terms, "n" is the number of terms, "a" is the first term, and "d" is the common difference.
Plugging in the values, we get:
S13 = (13/2)(2(20) + (13-1)(20)).
S13 = (13/2)(40 + 240).
S13 = 13(140).
S13 = 1820.
So, the sum of the terms from the 8th to the 20th (inclusive) is 1820.
2) Ah, let's ramble through this arithmetic progression with a spring in our step!
We have the AP 1 + 3 + 5 + ... with a common difference of 2. We want to find the least number of terms required to make a sum exceeding 4000.
Using the formula for the sum of an arithmetic progression, Sn = (n/2)(2a + (n-1)d), we can plug in the values:
4000 < (n/2)(2(1) + (n-1)(2)).
Simplifying, we get:
4000 < (n/2)(2 + 2n - 2).
4000 < (n/2)(2n).
8000 < n^2.
Therefore, we need a value of "n" greater than the square root of 8000, which is approximately 89.44.
Since we're talking about the number of terms, we round up to the nearest whole number, meaning we need at least 90 terms.