AP is a, a+d, a+2d, ..., a+(n-1)d where n is the number of terms, a is the first term and d the common difference.
The sum S=a+(a+d)+(a+2d)+...+(a+(n-1)d) which can be rearranged in pairs:
(a+a+(n-1)d)+(a+d+a+(n-2)d)+(a+2d+a+(n-3)d)+...
a+a+(n-1)d=2a+nd-d; a+d+a+(n-2)d=2a+d+nd-2d=2a+nd-d; a+2d+a+(n-3)d=2a+nd-d; ... Therefore, sum of each pair of terms is the same.
So there are n/2 multiples of the same number: 2a+nd-d, because there are n/2 pairs of sums of two terms. Sn=½n(2a+nd-d).
When n=20, S20=420=10(2a+19d), 2a+19d=42, 2a=42-19d, a=½(42-19d)=21-19d/2.
11th term=a+10d=21-19d/2+10d=21+d/2.
We're not told what d is. Let's see what happens if d=2: 11th term is 21+1=22.
a=½(42-38)=2, so the first 20 terms of the series are:
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40.
The series sums to 420 and the 11th term is 22.
If d=4, then a=½(42-76)=-17 and the 11th term is 21+2=23, but the series to 20 terms is:
-17, -13, -9, -5, -1, 3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 55, 59.
The series sums to 420 and the 11th term is 23. However the series starts with a negative number (which is OK, since an AP doesn't have to have positive terms).
If we assume a positive term increasing series then the 11th term is 22.
If d=-2 then we have a decreasing series starting with 40 and ending with 2, making the 11th term 20.