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Q.
The sixth term of an A.P. $a_1, a_2, a_3, \ldots . . . a_n$ is 2 . If the quantity $a_1 a_4 a_5$ is minimum then the common difference of the A.P., is
Sequences and Series
Solution:
$ a_6=a_1+5 d=2$
$\text { Now } P=a_1 a_4 a_5=a_1\left(a_1+3 d\right)\left(a_1+4 d\right)=(2-5 d)(2-2 d)(2-d)=2\left(4-16 d+17 d^2-5 d^3\right) $
$\text { Consider } f ( d )=-5 d ^3+17 d ^2-16 d +4 \text { [12th, 29-08-2010 P-1] } $
$\therefore f ^{\prime}( d )=-15 d ^2+34 d -16 \text {, for minima, } f ^{\prime}( d )=0 \Rightarrow d =\frac{2}{3}, \frac{8}{5} $
$f ^{\prime \prime}( d )=-30 d +34 $
$f ^{\prime \prime}\left(\frac{2}{3}\right)=-20+34>0 \Rightarrow d =\frac{2}{3} \text { gives minima} $