Question

Given f(3)=7 and f(5)=252 write the first 6 terms of the sequence. no decimals

Answer 1

$\bf \begin{array}{|cl|ll} \cline{1-2} term&value\\ \cline{1-2} f(1)&\\ f(2)&\\ f(3)&7\\ f(4)&7r\\ f(5)&7rr\\ &252\\ \cline{1-2} \end{array}\qquad \qquad 7rr=252\implies 7r^2=252\implies r^2=\cfrac{252}{7} \\\\\\ r^2=36\implies r=\sqrt{36}\implies r=6 \\\\[-0.35em] ~\dotfill\\\\ \begin{array}{|cl|ll} \cline{1-2} term&value\\ \cline{1-2} f(1)&\stackrel{\frac{7}{6}\div 6}{\cfrac{7}{36}}\\ &\\ f(2)&\stackrel{7\div 6}{\cfrac{7}{6}}\\ &\\ f(3)&7\\ f(4)&42\\ f(5)&252\\ f(6)&1512\\ \cline{1-2} \end{array}$

notice, once we know what the common factor "r" is, from the 3rd term we can simply multiply it by "r" to get the next term, and divide the 3rd term by "r" in order to get the previous term, namely the 2nd term, and then divide the 2nd by "r" to get the 1st one.