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Alisiya [41]
3 years ago
15

Can a sequence be both arithmetic and geometric?

Mathematics
1 answer:
artcher [175]3 years ago
4 0
An aritmetic sequence is like this
a_n=a_1+d(n-1) where a1=first term and d=common difference

geometric is a_n=a_1(r)^{n-1} where a1=first term and r=common ratio


can it be both aritmetic and geometric
hmm, that means that the starting terms should be the same

therfor we need to solve d(n-1)=(r)^{n-1}
what values of d and r make all natural numbers of n true?
are there values that make all natural numbers for n true?

when n=1, then d(1-1)=0 and r^(1-1)=1, so already they are not equal

the answer is no, a sequence cannot be both aritmetic and geometric
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(9\sin2x+9\cos2x)^2=81

Taking the square root of both sides gives two possible cases,

9\sin2x+9\cos2x=9\implies\sin2x+\cos2x=1

or

9\sin2x+9\cos2x=-9\implies\sin2x+\cos2x=-1

Recall that

\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta

If \alpha=2x and \beta=\dfrac\pi4, we have

\sin\left(2x+\dfrac\pi4\right)=\dfrac{\sin2x+\cos2x}{\sqrt2}

so in the equations above, we can write

\sin2x+\cos2x=\sqrt2\sin\left(2x+\dfrac\pi4\right)=\pm1

Then in the first case,

\sqrt2\sin\left(2x+\dfrac\pi4\right)=1\implies\sin\left(2x+\dfrac\pi4\right)=\dfrac1{\sqrt2}

\implies2x+\dfrac\pi4=\dfrac\pi4+2n\pi\text{ or }\dfrac{3\pi}4+2n\pi

(where n is any integer)

\implies2x=2n\pi\text{ or }\dfrac\pi2+2n\pi

\implies x=n\pi\text{ or }\dfrac\pi4+n\pi

and in the second,

\sqrt2\sin\left(2x+\dfrac\pi4\right)=-1\implies\sin\left(2x+\dfrac\pi4\right)=-\dfrac1{\sqrt2}

\implies2x+\dfrac\pi4=-\dfrac\pi4+2n\pi\text{ or }-\dfrac{3\pi}4+2n\pi

\implies2x=-\dfrac\pi2+2n\pi\text{ or }-\pi+2n\pi

\implies x=-\dfrac\pi4+n\pi\text{ or }-\dfrac\pi2+n\pi

Then the solutions that fall in the interval [0,2\pi) are

x=0,\dfrac\pi4,\dfrac\pi2,\dfrac{3\pi}4,\pi,\dfrac{5\pi}4,\dfrac{3\pi}2,\dfrac{7\pi}4

5 0
3 years ago
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alukav5142 [94]

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Step-by-step explanation:

ill suppose that you mean : "solve for b" not for d, since there is no d term in the formula given. So:

q = 4(a + b) \\ q = 4a + 4b \\ 4a + 4b = q \\ 4a + 4b - 4a = q - 4a \\ 4b = q - 4a \\ b =  \frac{q - 4a}{4}

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