How do you write an nth term rule for $r = 6$ $r=6$ and $a_{3} = 72$ $a_3=72$ ?

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How do you write an nth term rule for $r = 6$ $r=6$ and $a_{3} = 72$ $a_3=72$ ?

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Answer 1 · 2018-03-22T08:30:58

$U n = 2 \cdot 6^{n - 1}$ $Un= 2*6^(n-1)$

Explanation:

As we know that 72 is the third term and the common ratio is 6 so therefore we will divide 72 by 6 and then again by 6 in order to get the 1st term which is 2. Or in other terms if we do $6^{2}$ $6^2$ which is 36, then you divide 72 by 36 which will give you 2.

Then you insert the formula in the geometric series which is written above as the answer

Answer 2 · 2018-03-22T08:45:34

$a_{n} = 2 \times {(6)}^{n - 1}$ $a_n=2xx(color(red)(6))^(n-1)$

Explanation:

$the nth term of a geometric sequence is$ $"the nth term of a geometric sequence is"$

$∙ x a_{n} = a r^{n - 1}$ $•color(white)(x)a_n=ar^(n-1)$

$where a is the first term and r the common ratio$ $"where a is the first term and r the common ratio"$

$we are given r and require to find a$ $"we are given r and require to find a"$

$\Rightarrow a_{3} = a {(6)}^{2} = 72 \leftarrow n th term formula$ $rArra_3=a(6)^2=72larrcolor(blue)"n th term formula"$

$\Rightarrow 36 a = 72 \Rightarrow a = \frac{72}{36} = 2$ $rArr36a=72rArra=72/36=2$

$\Rightarrow a_{n} = 2 \times {(6)}^{n - 1} \leftarrow n^{t h} term rule$ $rArra_n=2xx(6)^(n-1)larrcolor(blue)n^(th) " term rule"$

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How do you write an nth term rule for $r = 6$ $r=6$ and $a_{3} = 72$ $a_3=72$ ?

2 Answers

Explanation:

Explanation:

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How do you write an nth term rule for r=6r=6r=6 and a_3=72a3=72a_3=72?

2 Answers

Explanation:

Explanation:

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How do you write an nth term rule for $r = 6$ $r=6$ and $a_{3} = 72$ $a_3=72$ ?