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  • In the expansion of  the 4th term is the greatest numerical value then y

In the expansion of (2+3x)^{10} the 4th term is the greatest numerical value then y lies in the intervals 

Option: 1

\left(2, \frac{64}{21}\right) \bigcup\left(\frac{64}{21}, \infty\right)


Option: 2

(0,1)


Option: 3

\\ \left(0, \frac{64}{21}\right) \\


Option: 4

\left(\frac{64}{21}, \infty\right)


Answers (1)

best_answer

Given 4th term is the numerically greatest term in the expansion \left(2+\frac{3 y}{8}\right)^{10}then

\left|t_3\right|<\left|t_4\right|$ and also $\left|t_5\right|<\left|t_4\right|\\

\left|\frac{t_3}{t_4}\right|<1 \text { and }\left|\frac{t_5}{t_4}\right|<1

Since, 

\frac{t_3}{t_4}=\frac{10 c_2\left(2^8\right)\left(\frac{3 y}{8}\right)^2}{10 c_3\left(2^7\right)\left(\frac{3 y}{8}\right)^3}   then

\begin{aligned} & \Rightarrow\left|\frac{t_3}{t_4}\right|=\left|\frac{2}{y}\right|<1 \\ & \Rightarrow \frac{2}{|y|}<1 \\ & \Rightarrow|y|>2 \end{aligned}

Since,

\frac{t_5}{t_4}=\frac{10 c_2\left(2^6\right)\left(\frac{3 y}{8}\right)^4}{10 c_3\left(2^7\right)\left(\frac{3 y}{8}\right)^3}   then

\begin{aligned} & \Rightarrow\left|\frac{t_5}{t_4}\right|=\left|\frac{21 y}{64}\right|<1 \\ & \Rightarrow \frac{21}{64}|y|<1 \\ & \Rightarrow|y|<\frac{64}{21} \end{aligned}

By combining both conditions, the interval of y is obtained as  \left(2, \frac{64}{21}\right) \bigcup\left(\frac{64}{21}, \infty\right)

Posted by

rishi.raj

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