a, b, and c are three integers such that a and b are less than 100, an

Not really. Look below. In DS questions, you need to absolutely sure that you will not get an answer by using the statements alone.

Sum of n terms of an arithmetic progression (a sequence in which each term after the 1st one is related to the previous term by a constant, k/11 in this case) =\(\frac{n*[2*a+(n-1)*d]}{2}\)

where, n =100, d = k/11

Substituting, we get,

Sum of 100 terms = 50* [2a+9k]

Per statement 2, after rearranging the terms, we get 2a+9k = 46. Thus this statement is sufficient to give the sum of 100 terms = 50*46 = a unique value.

Per statement 1, 2a+11k = 46 ---> a = (46-11k)/2 and as 'a' is a positive integer (>0), the only possible values of k to give integer value of 'a' are k = 4 or k =2

Thus we get , a=12 when k =2 or a=1 when k=4.

Case 2: a=1 when k=4 , NOT possible as we are told that 1 of 'a' and 'k' is a prime number and as both 1 and 4 are NOT primes, this case is discarded.

Case 1: a=12 when k =2 , possible as 2 is a prime number. Thus this is the only case possible and hence statement 1 is sufficient as well.

Thus both statements are sufficient on their own.

D is the correct answer.