### Binomial Theorem Question 23

#### Question 23 - 01 February - Shift 2

Let the sixth term in the binomial expansion of $(\sqrt{2^{\log _2}(10-3^{x})}+\sqrt[5]{2^{(x-2) \log _2 3}})^{m}$, in the increasing powers of $2^{(x-2) \log _2 3}$, be 21.

If the binomial coefficients of the second, third and fourth terms in the expansion are respectively the first, third and fifth terms of an A.P., then the sum of the squares of all possible values of $x$ is_____________

## Show Answer

#### Answer: (4)

#### Solution:

#### Formula: General term in the expansion, Condition for three no. in AP

$T_6={ }^{m} C_5(10-3^{x})^{\frac{m-5}{2}} \cdot(3^{x-2})=21$ ………(1)

${ }^{m} C_1,{ }^{m} C_2,{ }^{m} C_3$ are in A.P.

- ${ }^{m} C_2={ }^{m} C_1+{ }^{m} C_3$

Solving for $m$, we get

$m=2$ (rejected), 7

Put in equation (1)

$21 \cdot(10-3^{x}) \frac{3^{x}}{9}=21$

$3^{x}=3^{0}, 3^{2}$

$x=0,2$