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Mathematics 16 Online
OpenStudy (anonymous):

if the 1st term and common difference of G.P are each equal to x +\[\sqrt{5-x^2} i\] then the modulus of nterm

OpenStudy (anonymous):

@ghazi

OpenStudy (anonymous):

common difference of GP???

OpenStudy (anonymous):

is thatsupposed to be common ratio

OpenStudy (anonymous):

both are same \[\sqrt{5-x^2} i\]

OpenStudy (anonymous):

Is "modulus" here meaning "magnitude?"

OpenStudy (anonymous):

\[a _{n}=ar^{n-1}=(x+\sqrt{5-x^2}i)(x+\sqrt{5-x^2}i)^{n-1}\] \[=(x+\sqrt{5-x^2}i)^{n}\] if i is outside the square root modulus means\[\left| \right|\]

OpenStudy (anonymous):

Yup i is outside

OpenStudy (anonymous):

\[\left| (x+\sqrt{5-x^2}i)^n \right|\] not sure wat happens to i

OpenStudy (anonymous):

lol...i did nt get u

OpenStudy (anonymous):

from which step

OpenStudy (anonymous):

The answer shuld be 5^(n/2)

OpenStudy (anonymous):

well i said we have \[x(x)^{n-1}=x^n\] where x is that expression \[(x+\sqrt{5-x^2}i)^n\]

OpenStudy (anonymous):

ok... then

OpenStudy (anonymous):

\[i=\sqrt{-1}\] so \[(x+\sqrt{-1}\sqrt{5-x^2})^n\]

OpenStudy (anonymous):

which is \[(x+\sqrt{x^2-5})^n\]

OpenStudy (anonymous):

\[\left| (x+\sqrt{x^2-5})^n \right|\]

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