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2 years ago

5

WILL GIVE BRAINLIEST!!! How many terms are in the geometric sequence having a first term of 2, a last term of 32, and a common r

atio of -2?

1 answer:

Elza [17]2 years ago

7 0

Answer:

3

Step-by-step explanation:

first term a= 2

last term tn= 32

common ratio r= -2

number of terms

= tn=ar^n-1

32 = 2*(-2)^n-1

divide both sides by 2

32/2 = 2*(-2)^n-1/2

16 = (-2)^n-1

Then...use 16 in the power of 2

(-2)^4 = (-2)^n-1

The bases (-2), cancels out

so we have: 4 = n-1

collect like terms

4-1 = n

3 = n

n= 3

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2 years ago

Solve the following equations: (a) x^11=13 mod 35 (b) x^5=3 mod 64

tino4ka555 [31]

a.

$x^{11}=13\pmod{35}\implies\begin{cases}x^{11}\equiv13\equiv3\pmod5\\x^{11}\equiv13\equiv6\pmod7\end{cases}$

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$x^{11}\equiv (x^5)^2x\equiv x^3\equiv3\pmod5$

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b.

$x^5\equiv3\pmod{64}$

We have $\varphi(64)=32$ , so by Euler's theorem,

$x^{32}\equiv1\pmod{64}$

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13 = 3*4 + 1

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