Question

find the middle term in the expansion of
$( \frac{a}{y} - \frac{y}{a} ) {}^{12}$
?​

Answer 1

(4x² + 80x - 19200 = 0) ÷ 4

⇒ x² + 20x - 4800 = 0

⇒ x² + 80x - 60x - 4800 = 0

⇒ x(x + 80) - 60(x + 80) = 0

⇒ (x + 80) (x - 60) = 0

⇒ x = - 80 or x = 60

Speed cannot be negative. So, x = 60 is correct.

So, the speed of another train is 60 km/hr and speed of Deccan Queen is   60 + 20 = 80 km/hr

Answer.  

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Answer 2

$\large\underline{\sf{Solution-}}$

Given expansion is

$\rm :\longmapsto\: {\bigg(\dfrac{a}{y} - \dfrac{y}{a} \bigg) }^{12}$

We know,

Let us assume a binomial expansion,.

$\rm :\longmapsto\: {(x + a)}^{n}$

Now,

↝ Middle term in the binomial expansion depends upon the value of n.

• If n is even, there is only one middle term.

• If n is odd, there are 2 middle terms.

Here, n = 12 which is even.

Therefore, there is only one middle term and that term is given by [ (n/2) + 1 ] th term

So, 7th term is the middle term.

Now, We further know that

General Term of binomial expansion is given by

$\boxed{ \bf{ \:T_{r + 1} = \: ^nC_r \: {x}^{n - r} {a}^{r}}}$

So,

Middle term is

$\rm :\longmapsto\:T_{7}$

$\rm \: = \: \:T_{6 + 1}$

$\rm \: = \: \:^{12}C_6 \: {\bigg(\dfrac{a}{y} \bigg) }^{12 - 6}{\bigg( - \dfrac{y}{a} \bigg) }^{6}$

$\rm \: = \: \:^{12}C_6 \: {\bigg(\dfrac{a}{y} \bigg) }^{6}{\bigg( \dfrac{y}{a} \bigg) }^{6}$

$\rm \: = \: \:^{12}C_6$

$\rm \: = \: \:\dfrac{12!}{6! \: (12 - 6)!}$

$\rm \: = \: \:\dfrac{12!}{6! \: 6!}$

$\rm \: = \: \:\dfrac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6!}{6! \times \: 6 \times 5 \times 4 \times 3 \times 2}$

$\rm \: = \: \:\dfrac{12 \times 11 \times 10 \times 9 \times 8 \times 7 }{ \: 6 \times 5 \times 4 \times 3 \times 2}$

$\rm \: = \: \:924$

Hence,

Middle term in the expansion of

$\bf :\longmapsto\: {\bigg(\dfrac{a}{y} - \dfrac{y}{a} \bigg) }^{12}$

is

$\bf \: = \: \:924$

Additional Information :-

$\boxed{ \sf{ \:^nC_r = \dfrac{n!}{r! \: (n - r)!}}}$

$\boxed{ \sf{ \:^nC_r = \frac{n}{r} \: ^{n - 1}C_{r - 1}}}$

$\boxed{ \sf{ \:\dfrac{^nC_r}{^nC_{r - 1}} = \dfrac{n - r + 1}{r}}}$

$\boxed{ \sf{ \:^nC_r \: + \: ^nC_{r - 1} = ^{n + 1}C_r}}$