Question

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 AP, then the sum of the squares of all possible values of $x$ is

Answer 1

The correct answer is 4.
T6​=mC5​(10−3x)2m−5​⋅(3x−2)=21
mC1​,mC2​,mC3​ are in A.P.
2. mC2​=mC1​+mC3​
Solving for m, we get
m= 2(rejected), 7
Put in equation (1)
21.(10−3x)93x​=21
3x=30,32
x=0,2
Sum of the squares of all possible values of x=4