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The sixth term of an AP is 2 and its common difference is greater than 1
What is the common difference of the AP so that the product of the first, fourth and fifth terms is greatest? $\quad[2016-I I]$
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What is the common difference of the AP so that the product of the first, fourth and fifth terms is greatest? $\quad[2016-I I]$
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The correct answer is:
$8 / 5$
Let first term $=\mathrm{a} \&$ common difference $=\mathrm{x}$
$\therefore \quad a+5 x=2 \Rightarrow a=2-5 x$
$\begin{aligned} & \text { Let } \mathrm{P}=\mathrm{T}_{1} \times \mathrm{T}_{4} \times \mathrm{T}_{5} \\ \Rightarrow & \mathrm{P}=\mathrm{a}(\mathrm{a}+3 \mathrm{x})(\mathrm{a}+4 \mathrm{x}) \\ \Rightarrow & \mathrm{P}=(2-5 \mathrm{x})(2-5 \mathrm{x}+3 \mathrm{x})(2-5 \mathrm{x}+4 \mathrm{x}) \\ \Rightarrow & \mathrm{P}=-10 \mathrm{x}^{3}+34 \mathrm{x}^{2}-32 \mathrm{x}+8 \\ & \frac{d p}{d x}=0 \Rightarrow 15 \mathrm{x}^{2}-34 \mathrm{x}+16=0 \\ & \Rightarrow(5 \mathrm{x}-8)(3 \mathrm{x}-2)=0 \\ \Rightarrow & \mathrm{x}=\frac{8}{5},\left[\because x=\frac{2}{3} < 1\right] \end{aligned}$
$\therefore \quad a+5 x=2 \Rightarrow a=2-5 x$
$\begin{aligned} & \text { Let } \mathrm{P}=\mathrm{T}_{1} \times \mathrm{T}_{4} \times \mathrm{T}_{5} \\ \Rightarrow & \mathrm{P}=\mathrm{a}(\mathrm{a}+3 \mathrm{x})(\mathrm{a}+4 \mathrm{x}) \\ \Rightarrow & \mathrm{P}=(2-5 \mathrm{x})(2-5 \mathrm{x}+3 \mathrm{x})(2-5 \mathrm{x}+4 \mathrm{x}) \\ \Rightarrow & \mathrm{P}=-10 \mathrm{x}^{3}+34 \mathrm{x}^{2}-32 \mathrm{x}+8 \\ & \frac{d p}{d x}=0 \Rightarrow 15 \mathrm{x}^{2}-34 \mathrm{x}+16=0 \\ & \Rightarrow(5 \mathrm{x}-8)(3 \mathrm{x}-2)=0 \\ \Rightarrow & \mathrm{x}=\frac{8}{5},\left[\because x=\frac{2}{3} < 1\right] \end{aligned}$
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