How do you write a rule for the nth term of the geometric sequence given the two terms $a_{4} = 351, a_{7} = 13$ $a_4=351, a_7=13$ ?

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Answer 1 · 2017-05-17T19:50:19

$a_{n} = 9477 \cdot {(\frac{1}{3})}^{n - 1}$ $a_(n) = 9477 cdot (frac(1)(3))^(n - 1)$

We have: $a_{4} = 351$ $a_(4) = 351$ and $a_{7} = 13$ $a_(7) = 13$

The $n$ $n$ th term of a geometric sequence is given by:

$a_{n} = a_{1} r^{n - 1}$ $a_(n) = a_(1) r^(n - 1)$

Let's express the $4$ $4$ th and $7$ $7$ th terms using this rule:

$\Rightarrow a_{4} = 351$ $Rightarrow a_(4) = 351$

$\Rightarrow a_{1} r^{4 - 1} = 351$ $Rightarrow a_(1) r^(4 - 1) = 351$

$\Rightarrow a_{1} r^{3} = 351$ $Rightarrow a_(1) r^(3) = 351$ ----------- $(i)$ $(i)$

and

$\Rightarrow a_{7} = 13$ $Rightarrow a_(7) = 13$

$\Rightarrow a_{1} r^{7 - 1} = 13$ $Rightarrow a_(1) r^(7 - 1) = 13$

$\Rightarrow a_{1} r^{6} = 13$ $Rightarrow a_(1) r^(6) = 13$ ------------ $(i i)$ $(ii)$

Then, let's divide $(i i)$ $(ii)$ by $(i)$ $(i)$ :

$\Rightarrow \frac{a_{1} r^{6}}{a_{1} r^{3}} = \frac{13}{351}$ $Rightarrow frac(a_(1) r^(6))(a_(1) r^(3)) = frac(13)(351)$

$\Rightarrow r^{3} = \frac{1}{27}$ $Rightarrow r^(3) = frac(1)(27)$

$\Rightarrow r = \frac{1}{3}$ $Rightarrow r = frac(1)(3)$

Now, let's find the first term by substituting this value for the common ratio into $(i i)$ $(ii)$ :

$\Rightarrow a_{1} {(\frac{1}{3})}^{6} = 13$ $Rightarrow a_(1) (frac(1)(3))^(6) = 13$

$\Rightarrow \frac{a_{1}}{729} = 13$ $Rightarrow frac(a_(1))(729) = 13$

$\Rightarrow a_{1} = 9477$ $Rightarrow a_(1) = 9477$

Finally, let's substitute these values back into the rule for the $n$ $n$ th term:

$therefore a_(n) = 9477 cdot (frac(1)(3))^(n - 1)$

How do you write a rule for the nth term of the geometric sequence given the two terms a_4=351, a_7=13a4=351,a7=13a_4=351, a_7=13?