JAMB - Mathematics (2013 - No. 16)
The nth term of the progression \(\frac{4}{2}\), \(\frac{7}{3}\), \(\frac{10}{4}\), \(\frac{13}{5}\) is ...
\(\frac{1 - 3n}{n + 1}\)
\(\frac{3n + 1}{n + 1}\)
\(\frac{3n + 1}{n - 1}\)
\(\frac{3n - 1}{n + 1}\)
स्पष्टीकरण
Using Tn = \(\frac{3n + 1}{n + 1}\),
T1 = \(\frac{3(1) + 1}{(1) + 1} = \frac{4}{2}\)
T2 = \(\frac{3(2) + 1}{(2) + 1} = \frac{7}{3}\)
T3 = \(\frac{3(3) + 1}{(3) + 1} = \frac{10}{4}\)
T1 = \(\frac{3(1) + 1}{(1) + 1} = \frac{4}{2}\)
T2 = \(\frac{3(2) + 1}{(2) + 1} = \frac{7}{3}\)
T3 = \(\frac{3(3) + 1}{(3) + 1} = \frac{10}{4}\)