Question

Find the values of T1 and R for a geometric sequence with T6=4 and T10=324

Answer 1

Given:

A geometric sequence with sixth term '4' and tenth term '324'.

Required: First term and common ratio

Step-by-step explanation:

The general term of a geometric sequence is

`t_n=t_1r^(n-1)`

Since the sixth term is 4,

`\begin{gathered} t_6=4 \\ t_1r^5=4 \end{gathered}`

Since the tenth term is 324,

`\begin{gathered} t_(10)=324 \\ t_1r^9=324 \end{gathered}`

Dividing them.

`\begin{gathered} (t_1r^9)/(t_1r^5)=(324)/(4) \\ r^4=81 \\ \implies r=\pm3 \end{gathered}`

Substitute 3 of r into t6 =4.

`\begin{gathered} t_1\cdot3^5=4 \\ t_1=(4)/(243) \end{gathered}`

Substitute -3 of r into t6 =4.

`\begin{gathered} t_1\cdot(-3)^5=4 \\ t_1=-(4)/(243) \end{gathered}`

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:

`t_1=(4)/(243),r=3`