Find the value of k which satisfies 1 - 3 k- 1 - 4 k-1

The correct option is A 12 / 7 Given, k3+k4=1 ⇒4k+3k12=1 ⇒ 7k=12 ⇒ k=127 ...

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Elimination Method of Finding Solution of a Pair of Linear Equations

Find the valu...

Question

Find the value of $k$ which satisfies $\frac{1}{3} k + \frac{1}{4} k = 1$

\frac{1}{7}

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\frac{12}{7}

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\frac{7}{2}

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6

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12

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Solution

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α_{k} = cos \frac{k π}{7} + i sin \frac{k π}{7},

i = \sqrt{- 1}

\frac{\sum_{k = 1}^{12} | α_{k + 1} - α_{k} |}{\sum_{k = 1}^{3} | α_{4 k - 1} - α_{4 k - 2} |}

k

a_{k} = cos (\frac{k π}{7}) + i sin (\frac{k π}{7})

\frac{12 \sum k = 1 | a_{k + 1} - a_{k} |}{3 \sum k = 1 | a_{4 k - 1} - a_{4 k - 2} |}

k

(4 k - 1, k - 2), (k, k - 7)

(k - 1, - k - 2)

k

α_{k} = cos (\frac{k π}{7}) + i sin (\frac{k π}{7})

i = \sqrt{- 1}

\frac{12 \sum k = 1 | α_{k + 1} - α_{k} |}{3 \sum k = 1 | α_{4 k - 1} - α_{4 k - 2} |} =

S = \frac{1}{2 \times 7} + \frac{1}{7 \times 12} + \frac{1}{12 \times 17} + \cdot \cdot \cdot + \frac{1}{225 \times 257}

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