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Numerically, the greatest value of the term in the expansion of (3-5 x)^{15}, when x=1 / 5, is

Option: 1

3^{12}(455)


Option: 2

3^{11}(455)


Option: 3

3^{12}(91)


Option: 4

3^{11}(91)


Answers (1)

best_answer

(3-5 x)^{15}=3^{15}\left(1-\frac{5}{3} x\right)^{15}$.\\ By the formula of numerically greatest term,\\ $\frac{T_{r+1}}{T_r} \equiv \frac{n-r+1}{r}\left(\frac{5}{3} x\right) \geq 1$.\\ On putting $x=\frac{1}{5}$,\\ we obtain \\$ \frac{16-r}{r} \frac{5}{3} \cdot \frac{1}{5} \geq 1 $ Or $\quad 16-r \leq 3 r$ Or $r \leq 4$\\ $\Rightarrow \quad T_4$ and $T_5$ are numerically equal to each other and are greater than any other term. By calculations,\\ $ T_4={ }^{15} C_3 \cdot 3^{15}\left(-\frac{5}{3} \frac{1}{5}\right)^3\begin{aligned} & =\frac{15 !}{3 ! 12 !} 3^{12} \text { numerically } \\ & =3^{12} \cdot 35 \cdot 13=3^{12}(455) \\ T_5 & ={ }^{15} C_4 \cdot 3^{15}\left(-\frac{5}{3} \frac{1}{5}\right)^4=\frac{15 !}{4 ! 11 !} 3^{11} \\ & =3^{12}(455) \end{aligned}

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Devendra Khairwa

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