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 \(.........\).
- A \(6\)
- B \(4\)
- C \(8\)
- D \(2\)
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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