a)Sum of the squares of the first and second term is 116.
b)The seventh term is divisible by 10.
Answer - A
Explanation - let x be the first and y be the second term.
Hence 0 less than x less than y
From Statement 1: x^2 + y^2 = 116 only when x=2 and y=10.
Hence the first term is 4. This is the only combination that works - hence sufficient
Statement 2: let 10z be the 7th term (z is an integer)
Thus x+6(y-x)=10z => y=(10z+5x)/6. Hence the last digit of the numerator can be 5 or 0. Now y is also an integer => x must be even => least possible value of y =10.
Assuming different values of z we get
z=1; x=12 This is impossible (x should be less than y)
z=2; x=8; 8 10 12 14 16 18 20
z=3; x=6; 6 10 14 18 22 26 30 and so on -- hence insufficient
Hence the first term is 4. This is the only combination that works - hence sufficient
Statement 2: let 10z be the 7th term (z is an integer)
Thus x+6(y-x)=10z => y=(10z+5x)/6. Hence the last digit of the numerator can be 5 or 0. Now y is also an integer => x must be even => least possible value of y =10.
Assuming different values of z we get
z=1; x=12 This is impossible (x should be less than y)
z=2; x=8; 8 10 12 14 16 18 20
z=3; x=6; 6 10 14 18 22 26 30 and so on -- hence insufficient
