If (a_1=5) and the sum of the first (6) terms is (120), what is the common difference?

Correct answer: (6)

Direct answer: Option C, 6. For an arithmetic progression, the sum of the first n terms is \\(S_n=\\frac n2[2a+(n-1)d]\\). Here \\(a=5\\), \\(n=6\\), and \\(S_6=120\\). Substitute carefully: \\(120=\\frac62[2(5)+(6-1)d]\\). Thus \\(120=3(10+5d)=30+15d\\). Subtract 30: \\(90=15d\\). Divide by 15: \\(d=6\\). Option A, 4, would give sum \\(3(10+20)=90\\), not 120. Option B, 5, would give \\(3(10+25)=105\\). Option C, 6, gives \\(3(10+30)=120\\), so it is correct. Option D, 7, would give \\(3(10+35)=135\\). A direct check gives terms 5, 11, 17, 23, 29, 35, whose sum is 120. Remember: in the sum formula use n−1, not n, and solve the equation step by step.