Question

em uma PA,a10=34 e a17=68.calcule a32

Answer 1

$a_n=a_p+r(n-p)\\\\a_{17}=a_{10}+r(17-10)\\\\68=34+r(7)\\\\68-34=7r\\\\34=7r\\\\r=\dfrac{34}{7}$


$a_n=a_p+r(n-p)\\\\a_{32}=a_{17}+r(32-17)\\\\a_{32}=68+\dfrac{34}{7}(15)\\\\\\a_{32}=\dfrac{476}{7}+\dfrac{510}{7}\\\\\\\boxed{a_{32}=\dfrac{986}{7}}$