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JEE Mains · Maths · STD 11 - 7. binomial theoram

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

  1. A \(6\)
  2. B \(4\)
  3. C \(8\)
  4. D \(2\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(4\)

Step-by-step Solution

Detailed explanation

\(T _6={ }^{ m } C _{ o }\left(10-3^{ x }\right)^{\frac{ m -5}{2}} \cdot\left(3^{ x -2}\right)=21\) \({ }^{ m } C _1,{ }^{ m } C _2,{ }^{ m } C _3\) are in \(A.P.\) \(2.\) \({ }^{ m } C _2={ }^{ m } C _1+{ }^{ m } C _3\) Solving for \(m\), we get \(m =\) \(2\)(rejected), \(7\)…
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