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If [tex]t_n[/tex] = 11nāˆ’6, find [tex]t_1[/tex], [tex]t_1_0[/tex] and [tex]t_n_+_1-t_n[/tex]

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[tex]\displaystyle\bf\\t_n=11n-6\\\\t_1=11\cdot1-6=11-6=\boxed{\bf5}\\\\t_{10}=11\cdot10-6=110-6=\boxed{\bf104}\\\\t_{n+1}-t_n=11(n+1)-6-(11\cdot n-6)=11n+11-6-11n+6=\boxed{\bf11}[/tex]

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