AP EAMCET · Maths · Binomial Theorem
If \(x\) is positive real number and the first negative term in the expansion of \((1+\mathrm{x})^{27 / 5}\) is \(\mathrm{t}_{\mathrm{k}}\) then \(\mathrm{k}=\)
- A \(5\)
- B \(6\)
- C \(7\)
- D \(8\)
Answer & Solution
Correct Answer
(D) \(8\)
Step-by-step Solution
Detailed explanation
\(n = 27/5 = 5.4\) For the coefficient to be negative, the factor \((n-r+1)\) must be the first to become negative. \(n-r+1 \(5.4-r+1 \(6.4-r \(r > 6.4\) The smallest integer \(r\) is \(7\). The term is \(t_k = T_{r+1}\). \(k = r+1 = 7+1 = 8\)
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