Question

А10=3, а13=5, найти а28

Answer 1

$d= \dfrac{a_{13}-a_{10}}{13-10} = \dfrac{5-3}{3} = \dfrac{2}{3}$

      $a_{10}=a_1+9d\,\,\,\, \Rightarrow\,\,\,\,\, a_1=a_{10}-9d=3-9\cdot\dfrac{2}{3} =-3$

$a_{28}=a_1+27d=-3+27\cdot\dfrac{2}{3} =15$