Find the (10)th term of the AP (\frac{1}{2},1,\frac{3}{2},2,\ldots).

Correct answer: (5)

The direct answer is option C: the 10th term is \(5\). An AP has a first term \(a\), a common difference \(d\), and nth term \(a_n=a+(n-1)d\). From \(\frac12,1,\frac32,2,\ldots\), the first term is \(a=\frac12\). Each term increases by \(\frac12\), so \(d=\frac12\). For the 10th term, \(a_{10}=\frac12+(10-1)\frac12=\frac12+\frac92=\frac{10}{2}=5\). Option A, 4, is the 8th term, because \(\frac12+7\cdot\frac12=4\). Option B, \(\frac92\), is the 9th term. Option C, 5, is exactly the 10th term. Option D, \(\frac{11}{2}\), is the 11th term. A useful check is to notice that the numerator sequence is 1, 2, 3, 4, so the 10th term has numerator 10 over 2. Memory cue: use \(a_n=a+(n-1)d\), not \(a+nd\).