Step 1: Write the middle four terms in terms of a and d.
Let the first term be a and the common difference be d. With 20 terms, the terms sit symmetrically around the centre between the 10th and 11th terms, so the four middle terms are the 9th, 10th, 11th and 12th terms.
These are \(a+8d, a+9d, a+10d, a+11d\), and their sum is \(4a+38d\). We are told this sum is \(-22\), so \(4a+38d=-22\). This equation is already available from the question itself, but it has two unknowns, a and d, so one more independent equation is needed.
Step 2: Check statement 1 alone.
Statement 1 gives the sum of the first four terms: \(a+(a+d)+(a+2d)+(a+3d) = 4a+6d = 74\).
Now we have two equations: \(4a+38d=-22\) and \(4a+6d=74\). Subtracting the second from the first: \(32d=-96\), so \(d=-3\).
Substitute back: \(4a+6(-3)=74 \Rightarrow 4a-18=74 \Rightarrow 4a=92 \Rightarrow a=23\).
Both a and d come out as single, definite numbers, so statement 1 alone is sufficient.
Step 3: Check statement 2 alone.
Statement 2 says the difference between the first term and common difference is 26, i.e. \(a-d=26\), so \(a=26+d\).
Substitute into \(4a+38d=-22\): \(4(26+d)+38d=-22 \Rightarrow 104+4d+38d=-22 \Rightarrow 42d=-126 \Rightarrow d=-3\).
Then \(a=26+(-3)=23\), the same values as before. Again both a and d come out as single numbers, so statement 2 alone is also sufficient.
Final Answer:
Each statement, combined with the sum given in the question, independently pins down a = 23 and d = -3, so either statement alone is sufficient.
\[ \boxed{\text{d - either statement alone is sufficient, a = 23, d = -3}} \]