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3 years ago

A recursive rule for a geometric sequence is a1=3;an=12an−1 . What is the explicit rule for this sequence?

1 answer:

6 0

From the recursive rule, you can tell the initial value is 3 and the common ratio is 12. The explicit rule is always
.. (n-th term) = (initial value)*(common ratio)^(n -1)

Subsituting the values you know, you have
.. an = 3*12^(n-1)

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3 years ago

a father is now 38 years older than his son. Ten years ago he was twice as old as his son. let x represent the age of the son no

Helga [31]

The present age of father is 86 years old and present age of son is 48 years old

Solution:

Given that, a father is now 38 years older than his son

Ten years ago he was twice as old as his son

Let "x" be the age of son now

Therefore, from given,

Father age now = 38 + age of son now

Father age now = 38 + x

Ten years ago he was twice as old as his son

Age of son ten years ago = age of son now - 10

Age of son ten years ago = x - 10

Age of father ten years ago = 38 + x - 10

Then we get,

Age of father ten years ago = twice the age of son ten years ago

38 + x - 10 = 2(x - 10)

28 + x = 2x - 20

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x = 48

Thus son age now is 48 years old

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Thus present age of father is 86 years old and present age of son is 48 years old

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3 years ago

Prove that (Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A

iVinArrow [24]

Answer:

The answer is below

Step-by-step explanation:

We need to prove that:

(Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A.

Firstly, 1 / cos A = sec A, 1 / sin A = cosec A and tanA = sinA / cosA.

Also, 1 + tan²A = sec²A; sec²A - 1 = tan²A

$\frac{\sqrt{secA-1} }{\sqrt{secA+1} } +\frac{\sqrt{secA+1} }{\sqrt{secA-1} } =\frac{(\sqrt{secA-1)}(\sqrt{secA-1})+(\sqrt{secA+1)}(\sqrt{secA+1}) }{(\sqrt{secA+1})(\sqrt{secA-1}) } \\\\=\frac{secA-1+(secA+1)}{\sqrt{sec^2A-secA+secA-1} } \\\\=\frac{2secA}{\sqrt{sec^2A-1} } \\\\=\frac{2secA}{\sqrt{tan^2A} } \\\\=\frac{2secA}{tanA} \\\\=\frac{2*\frac{1}{cosA} }{\frac{sinA}{cosA} }\\\\= 2*\frac{1}{cosA}*\frac{cosA}{sinA}\\\\=2*\frac{1}{sinAA}\\\\=2cosecA$

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3 years ago