JEE Main 2014 · 12 Apr · Shift 1 · Mathematics Q67

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MathematicsSingle correct+4 / -1Question 67 of 90

If $\left(2+\dfrac x3\right)^{55}$ is expanded in the ascending powers of $x$ and the coefficients of powers of $x$ in two consecutive terms of the expansion are equal, then these terms are :

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and 8

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and 9

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and 29

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and 28

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Answer and worked solution

Official answer: B

$T_{r+1}=\binom{55}r2^{55-r}\left(\frac x3\right)^r$ . Equal consecutive coefficients:
$\frac{T_{r+1}}{T_r}=\frac{56-r}r\cdot\frac1{2\cdot3}=1\implies r=8.$
So $T_8$ and $T_9$ : the 8th and 9th terms.

Topics:General and middle terms

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JEE Main 2014 · 12 Apr · Shift 1, Mathematics question 67