Question

find the middle term in the expansion of( x/a-a/x)^10​

Answer 1

In the expansion of (a+b)n, there are n+1 terms.Therefore, in the expansion of (x3+9y)10, there are 11 terms.Therefore, 6th term will be middle term6th term=10C5(x3)5(9y)5=10×9×8×7×61×2×3×4×5×x5243×81×243×y5=9×4×7×x51×81×y5=20412x5y5

Answer 2

SOLUTION:-

Given:

The expansion of;

$( \frac{x}{a} - \frac{a}{x} ) {}^{10}$

To find:

The middle term.

Explanation:

We have, n=10 [Even number]

We know that, formula of the middle term;

$= > ( \frac{n}{2} + 1) {}^{th}$

Therefore,

$= > (\frac{10}{2} + 1) {}^{th} \\ \\ = > ( \frac{10 + 2}{2} ) {}^{th} \\ \\ = > ( \frac{12}{2} ) {}^{th} \\ \\ = > 6th \:term$

Now,

${}^{T} 6 = {}^{T} 5 + 1 = {}^{10} C5( \frac{x}{a} ) {}^{10 - 5} ( \frac{ - a}{x} ) {}^{5} \\ \\ = > - {}^{10} C5 = ( { \frac{x}{a}) }^{5} ( { \frac{a}{x}) }^{5} \\ \\ = > - \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5}{5 \times 5 \times 4 \times 3 \times 2 \times 1} ( { \frac{x}{a} )}^{5} ( \frac{x}{a} ) {}^{ - 5} \\ \\ = > - \frac{30240}{120} \\ \\ = > - 252$

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