In an arithmetic sequence, (a_6=31) and the common difference is (4). What is (a_n)?

(4n+7)

For an arithmetic sequence, the nth term is found from \\(a_n=a_1+(n-1)d\\), where \\(a_1\\) is the first term and \\(d\\) is the common difference. We are given \\(a_6=31\\) and \\(d=4\\). Substituting the sixth position gives \\(31=a_1+5(4)\\), so \\(a_1=31-20=11\\). Therefore, \\(a_n=11+(n-1)4=11+4n-4=4n+7\\). Hence option A is correct.

The check at the given position confirms the result: putting \\(n=6\\) into \\(4n+7\\) gives \\(4(6)+7=31\\), exactly as required. The expression \\(31+4n\\) incorrectly treats 31 as the first term and gives 55 at the sixth position. The other linear options also fail to preserve the stated sixth term or common difference. Thus the supplied answer A is mathematically consistent and the reasoning uses the correct relation for an arithmetic progression.

This is a Class 9 Mathematics question. Chapter: Sequences and Progressions. Topic: nth term.