Given:
a₁ = 2/5
a₂ = 4/5
Since the first 2 terms are given, let's determine the remaining 3 terms using the given formula:
[tex]\text{ a}_n\text{ = a}_{n-2}\text{ }.\text{ a}_{n-1}[/tex]We get,
For a₃,
[tex]\text{ a}_3\text{ = a}_{3-1}\text{ }.\text{ a}_{3-2}\text{ = a}_1\text{ }.\text{ a}_2[/tex][tex]\text{ a}_3\text{ = \lparen}\frac{2}{5})(\frac{4}{5})\text{ = }\frac{8}{25}[/tex]For a₄,
[tex]\text{ a}_4\text{ = a}_{4-2}\text{ }.\text{ a}_{4-1}\text{ = a}_2\text{ }.\text{ a}_3[/tex][tex]\text{ a}_4\text{ = \lparen}\frac{4}{5})(\frac{8}{25})\text{ = }\frac{20}{125}\text{ = }\frac{4}{25}[/tex]For a₅,
[tex]\text{ a}_5\text{ = a}_{5-2}\text{ }.\text{ a}_{5-1}\text{ = a}_3\text{ }.\text{ a}_4[/tex][tex]\text{ a}_5\text{ = \lparen}\frac{8}{25})(\frac{4}{25})\text{ = }\frac{32}{625}[/tex]Therefore, the first five terms of the sequence are the following:
[tex]\text{ }\frac{2}{5}\text{ , }\frac{4}{5}\text{ , }\frac{8}{25}\text{ , }\frac{4}{25}\text{ , }\frac{32}{625}[/tex]For another two terms.
For a₆,
[tex]\text{ a}_6\text{ = a}_{6-2}\text{ }.\text{ a}_{6-1}\text{ = a}_4\text{ }.\text{ a}_5[/tex][tex]\text{ a}_6\text{ = \lparen}\frac{4}{25})(\frac{32}{625})\text{ = }\frac{128}{15,625}[/tex]For a₇,
[tex]\text{ a}_7\text{ = a}_{7-2}\text{ }.\text{ a}_{7-1}\text{ = a}_5\text{ }.\text{ a}_6[/tex][tex]\text{ a}_7\text{ = \lparen}\frac{32}{625})(\frac{128}{15,625})\text{ = }\frac{4,096}{9,765,625}[/tex]