Find the 7th term in the expansion of \[\left( \frac{4x}{5} + \frac{5}{2x} \right)^8\]

We need to find the 7th term in the given expression. \[\because T_7 = T_{6 + 1}\] \[\therefore T_7 = T_{6 + 1} \]\[ =^{8}{}{C}_6 \left( \frac{4x}{5} \right)^{8 - 6} \left( \frac{5}{2x} \right)^6 \]\[ = \frac{8 \times 7 \times 4 \times 4 \times 125 \times 125}{2 \times 1 \times 25 \times 64} x^2 \left( \frac{1}{x^6} \right)\]\[ = \frac{4375}{x^4}\]

Find the 7th Term in the Expansion of ( 4 X 5 + 5 2 X ) 8 - Mathematics

प्रश्न

Find the 7th term in the expansion of ${(\frac{4 x}{5} + \frac{5}{2 x})}^{8}$

उत्तर

We need to find the 7th term in the given expression.

∵ T_{7} = T_{6 + 1}

$∴ T_{7} = T_{6 + 1}$
$=^{8} C_{6} {(\frac{4 x}{5})}^{8 - 6} {(\frac{5}{2 x})}^{6}$
$= \frac{8 \times 7 \times 4 \times 4 \times 125 \times 125}{2 \times 1 \times 25 \times 64} x^{2} (\frac{1}{x^{6}})$
$= \frac{4375}{x^{4}}$

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Rth Term from End

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पाठ 18: Binomial Theorem - Exercise 18.2 [पृष्ठ ३७]

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पाठ 18 Binomial Theorem
Exercise 18.2 | Q 5 | पृष्ठ ३७

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Find the 7th Term in the Expansion of ( 4 X 5 + 5 2 X ) 8 - Mathematics

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Find the 7th Term in the Expansion of ( 4 X 5 + 5 2 X ) 8 - Mathematics

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