The nth term of the progression \(\frac{4}{2}\), \(\frac{7}{3}\), \(\frac{10}{4}\), \(\frac{13}{5}\) is …

Explanation

Using T_n = \(\frac{3n + 1}{n + 1}\),

T₁ = \(\frac{3(1) + 1}{(1) + 1} = \frac{4}{2}\)

T₂ = \(\frac{3(2) + 1}{(2) + 1} = \frac{7}{3}\)

T₃ = \(\frac{3(3) + 1}{(3) + 1} = \frac{10}{4}\)

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