Question

Which are the first five terms of a geometric sequence in which the second term is 6 and the fourth term is 54?

Answer 1

now, let's recall that a geometric sequence is one that uses some "r" common ratio to get the next term, by simply multiplying the current term by it.

$\bf \begin{array}{|cl|ll} \cline{1-2} term&value\\ \cline{1-2} a_1&\underline{\qquad }\\&\\ a_2&6\\&\\ a_3&\underline{6(r)}\\&\\ a_4&6(r)(r)\\&\\ &54\\ \cline{1-2} \end{array}\qquad \implies \begin{array}{llll} 54=6r^2\implies \cfrac{54}{6}=r^2\implies 9=r^2\\\\ \sqrt{9}=r\implies 3=r \end{array} \\\\[-0.35em] ~\dotfill\\\\ a_1=6\div 3\implies a_1=2~\hfill a_3=6(3)\implies a_3=18$

and of course, the next term or a₅ = 54(3) --> a₅ = 162.