Question

If $'a'$ is the middle term in the expansion of $(2x - 3y)^8$ and $b, c$ are the middle terms in the expansion of $(3x + 4y)^7$ , then the value of $\frac{b +c}{a}$ ,when $x = 2$ and $y = 3$, is

• A. $\frac{1}{2}$
• B. $\frac{2}{3}$
• C. $1$
• D. $2$

Answer 1

Answer: (D)

$2$

Answer 2

In the expansion of $(2 x-3 y)^{8}$ middle term is $5$ th term.
$\Rightarrow a ={ }^{8} C_{4}(2 x)^{4}(3 y)^{4}(-1)^{4}$
$=\frac{8 !}{4 ! 4 !} \times 2^{4} \times 3^{4} \times x^{4} y^{4}$
$=70 \times 2^{4} \times 3^{4} \times 2^{4} \times 3^{4}(\because x=2, y=3)$
Similarly, in the expansion of $(3 x+4 y)^{7}, 4$ th and $5$ th terms are middle terms.
$\therefore b={ }^{7} C_{3}(3 x)^{4}(4 y)^{3}=35 \times 3^{7} \times 2^{10}$
and $c={ }^{7} C_{4}(3 x)^{3}(4 y)^{4}=35 \times 3^{7} \times 2^{11}$
Now, $b + c = 35 \times 3^7 \times 2^{10} + 35 \times 3^7 \times 2^{11}$
$= 35 \times 3^7 \times 2^{10} (1 + 2) = 35 \times 3^8 \times 2^{10}$
$\therefore \frac{b + c}{a} = \frac{35 \times 3^8 \times 2^{10}}{70 \times 2^8 \times 3^8} = \frac{2^{10}}{2^9} = 2$