In a geometrical progression first term and common ratio are both 12-3 - i- Then the absolute value of the nth term of the progression is

The correct option is C 1 α = 12(√(3) + i) ,r = 12(√(3) + i) ∴ T_n = α r^n 1 = [ 12(√(3) + i) ]^n = (r e^i θ)^n whe ...

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Standard XII

Mathematics

Complex Numbers

In a geometri...

Question

In a geometrical progression first term and common ratio are both $\frac{1}{2} (\sqrt{3} + i)$ . Then the absolute value of the nth term of the progression is

2^{n}

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4^{n}

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none of these

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Solution

The correct option is C 1
$α = \frac{1}{2} (\sqrt{3} + i), r = \frac{1}{2} (\sqrt{3} + i)$
$∴ T_{n} = α r^{n - 1} = {[\frac{1}{2} (\sqrt{3} + i)]}^{n} = (r e^{i θ})^{n}$
where r = 1 and $θ = \frac{π}{6}$
$∴ T_{n} = e^{i n θ} ∴ | T_{n} | = | e^{i n θ} | = 1$

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In a geometrical progression first term and common ratio are both 12(√3+i). Then the absolute value of the nth term of the progression is

The correct option is C 1α=12(√3+i),r=12(√3+i)∴Tn=αrn−1=[12(√3+i)]n=(reiθ)nwhere r = 1 and θ=π6∴Tn=einθ∴|Tn|=|einθ|=1

In a geometrical progression first term and common ratio are both $\frac{1}{2} (\sqrt{3} + i)$ . Then the absolute value of the nth term of the progression is

The correct option is C 1
$α = \frac{1}{2} (\sqrt{3} + i), r = \frac{1}{2} (\sqrt{3} + i)$
$∴ T_{n} = α r^{n - 1} = {[\frac{1}{2} (\sqrt{3} + i)]}^{n} = (r e^{i θ})^{n}$
where r = 1 and $θ = \frac{π}{6}$
$∴ T_{n} = e^{i n θ} ∴ | T_{n} | = | e^{i n θ} | = 1$