Given:
A geometric sequence with sixth term '4' and tenth term '324'.
Required: First term and common ratio
Explanation:
The general term of a geometric sequence is
[tex]t_n=t_1r^{n-1}[/tex]
Since the sixth term is 4,
[tex]\begin{gathered} t_6=4 \\ t_1r^5=4 \end{gathered}[/tex]
Since the tenth term is 324,
[tex]\begin{gathered} t_{10}=324 \\ t_1r^9=324 \end{gathered}[/tex]
Dividing them.
[tex]\begin{gathered} \frac{t_1r^9}{t_1r^5}=\frac{324}{4} \\ r^4=81 \\ \implies r=\pm3 \end{gathered}[/tex]
Substitute 3 of r into t6 =4.
[tex]\begin{gathered} t_1\cdot3^5=4 \\ t_1=\frac{4}{243} \end{gathered}[/tex]
Substitute -3 of r into t6 =4.
[tex]\begin{gathered} t_1\cdot(-3)^5=4 \\ t_1=-\frac{4}{243} \end{gathered}[/tex]
There are two possible geometric sequences:
1) First term, t1 = -4/243 and common ratio, r = -3
2) First term, t1 = 4/243 and common ratio, r = 3.
The correct option is option (b)
Final Answer:
[tex]t_1=\frac{4}{243},r=3[/tex]
