\(21q^2x^5\)
\(21q^4x^3\)
\(35q^3x^4\)
\(35q^5x^2\)
Correct answer is C
rth term of a binomial expansion = \(^nC_r - _1a^{n - (r - 1)} b^{r - 1}\)
n = 7, r = 5 ∴ r - 1 = 4
5th term = \(^7C_4 q^{7 - 4} x^4\)
= \(^7C_4 q^3x^4\)
\(\therefore 35q^3x^4\)
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