JEE Main 2022 · 27 Jul · Shift 2 · Mathematics Q23

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MathematicsInteger+4 / -1Question 23 of 90

Let for the 9 $^{\text{th}}$ term in the binomial expansion of $(3+6x)^n$ , in the increasing powers of $6x$ , to be the greatest for $x=\dfrac32$ , the least value of n is $n_0$ . If k is the ratio of the coefficient of $x^6$ to the coefficient of $x^3$ , then $k+n_0$ is equal to :

Your answer

Answer and worked solution

Official answer: 24

Greatest term. $\frac{T_{r+1}}{T_r}=\frac{3(n-r+1)}r$ at $x=\frac32$ . $T_9\ge T_8$ needs $n\ge10$ , and $T_9\ge T_{10}$ needs $n\le11$ . So $n_0=10$ .

Ratio. $k=\frac{\binom{10}6\,3^4\,6^6}{\binom{10}3\,3^7\,6^3}=14$ .

Answer: 24

Topics:General and middle terms Binomial expansion

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JEE Main 2022 · 27 Jul · Shift 2, Mathematics question 23