[tex] \boxed {\begin{array}{lclcl} \sf {First \: Eigth \: Term : } \\ \\ - 4, - 3, - 2, - 1, \:0,\: 1, \:2, \: \rm {and} \: 3\end{array}}[/tex]
Apply the general formula that has mentioned:
[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\begin{array}{l} \sf \large{ an} = \frac{ \large n {}^{2} - 25}{ \large n + 5} \end{array}}[/tex]
1. First Term; n = 1
[tex] \sf \small {a_1 } \normalsize= \frac{n {}^{2} - 25}{n + 5} = \frac{1 {}^{2} - 25 }{1 + 5} = \frac{1 - 25}{6} = \frac{ - 24 \: \: }{6} = \small- 4[/tex]
2. Second Term; n = 2
[tex] \sf \small {a_2} = \normalsize \frac{n {}^{2} - 25}{n + 5} = \frac{2 {}^{2} - 25 }{2 + 5} = \frac{4 - 25}{7} = \frac{ - 21 \: \: }{7} = \small- 3[/tex]
3. Third Term; n = 3
[tex] \sf \small {a_3 }= \normalsize \frac{n {}^{2} - 25}{n + 5} = \frac{3 {}^{2} - 25 }{3 + 5} = \frac{9 - 25}{8} = \frac{ - 16 \: \: }{8} = \small- 2[/tex]
4. Fourth Term; n = 4
[tex] \sf \small {a_4} = \normalsize \frac{n {}^{2} - 25}{n + 5} = \frac{4 {}^{2} - 25 }{4 + 5} = \frac{16 - 25}{9} = \frac{ - 9 \: \: }{9} = \small-1[/tex]
5. Fifth Term; n = 5
[tex] \sf \small {a_5} = \normalsize \frac{n {}^{2} - 25}{n + 5} = \frac{5 {}^{2} - 25 }{5 + 5} = \frac{25 - 25}{10} = \frac{ 0 }{9} = \small0[/tex]
6. Sixth Term; n = 6
[tex] \sf \small {a_6} = \normalsize\frac{n {}^{2} - 25}{n + 5} = \frac{6 {}^{2} - 25 }{6 + 5} = \frac{36 - 25}{11} = \frac{ 11 }{11} = \small1[/tex]
7. Seventh Term; n = 7
[tex] \sf \small{ a_7 }= \normalsize \frac{n {}^{2} - 25}{n + 5} = \frac{7 {}^{2} - 25 }{7 + 5} = \frac{49 - 25}{12} = \frac{ 24}{12} = \small2[/tex]
8. Eigth Term; n = 8
[tex] \sf \small {a_8} = \normalsize \frac{n {}^{2} - 25}{n + 5} = \frac{8 {}^{2} - 25 }{8+ 5} = \frac{64 - 25}{13} = \frac{ 39}{13} = \small3[/tex]
