A) \[x=\frac{8}{5}\]
B) \[x=\frac{5}{4}\]
C) \[x=2/3\]
D) None of these
Correct Answer: C
Solution :
Let \[a\] be the first term and \[x\] be the common difference of the A.P. Then \[a+5x=2\]\[\Rightarrow \]\[a=2-5x\] Let \[P={{a}_{1}}{{a}_{4}}{{a}_{5}}=a\,(a+3x)\,(a+4x)\] \[=(2-5x)(2-2x)(2-x)=2(-5{{x}^{3}}+17{{x}^{2}}-16x+4)\] Now\[\frac{dP}{dx}=0\]\[\Rightarrow \]\[x=\frac{8}{5},\ \frac{2}{3}\]. Clearly, \[\frac{{{d}^{2}}P}{d{{x}^{2}}}>0\] for \[x=\frac{2}{3}\] Hence \[P\] is least for\[x=\frac{2}{3}\].
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