Find the (7)th term of the AP (\frac{2}{3},\frac{5}{3},\frac{8}{3},\frac{11}{3},\ldots).

Correct answer: (\frac{20}{3})

The direct answer is option C: the 7th term is \(\frac{20}{3}\). The terms are \(\frac23,\frac53,\frac83,\frac{11}{3},\ldots\). Subtract consecutive terms: \(\frac53-\frac23=1\), \(\frac83-\frac53=1\), so the common difference is \(d=1\). The first term is \(a=\frac23\). Using \(a_n=a+(n-1)d\), we get \(a_7=\frac23+(7-1)(1)=\frac23+6\). Write 6 with denominator 3: \(6=\frac{18}{3}\). Therefore \(a_7=\frac23+\frac{18}{3}=\frac{20}{3}\). Option A, \(\frac{18}{3}=6\), misses the first term’s \(\frac23\). Option B, \(\frac{19}{3}\), is one third too small. Option C matches the calculation. Option D, \(\frac{21}{3}=7\), is one third too large. Another check is that the numerators increase by 3: 2, 5, 8, 11, 14, 17, 20; hence the seventh term is \(20/3\). Memory cue: count six jumps from the first term to reach the seventh term.