A) 7th term
B) 5th term
C) 8th term
D) 6th term
Correct Answer: C
Solution :
Since, (r + 1)th term in the expansion of \[{{(1+x)}^{27/5}}\] \[=\frac{\frac{27}{5}\left( \frac{27}{5}-1 \right).....\left( \frac{27}{5}-r+1 \right)}{r!}{{x}^{r}}\] Now, this term will be negative, if the last factor in numerator is the only one negative factor. \[\Rightarrow \,\,\frac{27}{5}-r+1<0\Rightarrow \frac{32}{5}<r\] \[\Rightarrow \] \[6.4<r\Rightarrow \] least value of r is 7. Thus, first negative term will be 8th.
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