In an AP, (a_{10}=31) and (a_{18}=55). What is (a_6)?

Correct answer: (19)

The general term of an arithmetic progression is \(a_n=a+(n-1)d\), and consecutive terms differ by the constant common difference d. The indices 10 and 18 are eight places apart. Their term values differ by 55−31=24, so \(8d=24\), giving \(d=3\). Moving from the tenth term back to the sixth term means moving four positions backward.

Each backward move subtracts the common difference. Therefore, \(a_6=a_{10}-4d=31-4(3)=31-12=19\). Choice B is correct. A common error is to use five steps instead of four; the index difference is \(10−6=4\), not the value of either index. The result is also consistent because adding 3 twelve times from the sixth term reaches the eighteenth term.