Question:
Find the 17th term in the following sequence whose nth term is $\mathrm{s} \mathrm{a}_{\mathrm{n}}=4 \mathrm{n}-3 ; \mathrm{a}_{17}, \mathrm{a}_{24}$
Solution:
Substituting n = 17, we obtain
$a_{17}=4(17)-3=68-3=65$
Substituting n = 24, we obtain
$\mathrm{a}_{24}=4(24)-3=96-3=93$
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