<img src="https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B0.0001%7D%20-%5Csqrt%5B5%5D%7B0.00032%7D" id="TexFormula1" title="\sqrt[4]{0.

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uranmaximum [27]

2 years ago

0001} -\sqrt[5]{0.00032}" alt="\sqrt[4]{0.0001} -\sqrt[5]{0.00032}" align="absmiddle" class="latex-formula">

Mathematics

1 answer:

Dmitry_Shevchenko [17]2 years ago

6 0

Hi!-0.041511 is the answer.

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a sequence has a first number of 3 and a half common difference of 4 what is the recursive rule of nth term

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Answer:

$f(n) = f(n-1) + 8$ for $n > 1$

Step-by-step explanation:

Given

$f(1) = 3$ -- First Term

$\frac{1}{2}d = 4$ --- half common difference

Required

Find the recursive rule

First, we calculate the common difference

$\frac{1}{2}d = 4$

Multiply through by 2

$2 * \frac{1}{2}d = 2 * 4$

$d = 8$

The second term of the sequence is:

$f(2) = 3 + 8 = 11$

The third term is:

$f(3) = 11 + 8 = 20$

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$f(1) = 3$

$f(2) = 3 + 8$

Substitute f(1) for 3

$f(2) = f(1) + 8$

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$f(2) = f(2-1) + 8$

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$f(n) = f(n-1) + 8$

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$f(3) = 11 + 8$

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$f(3) = f(2) + 8$

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$f(3) = f(3-1) + 8$

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$f(n) = f(n-1) + 8$

Hence, the recursive is:

$f(n) = f(n-1) + 8$ for $n > 1$

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3 years ago