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Q.
The sixth term of an A.P. is equal to $2 .$ The value of the common difference of the A.P. which makes the product $a_{1}\, a_{4}\, a_{5}$ greatest, is
Sequences and Series
Solution:
Given: $a_{1}+5 d=2$
Let $ y =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) $
$\left(\right.$ Putting $\left.a_{1}=2-5 d\right)$
$=2\left(4-16 d+17 d^{2}-5 d^{3}\right)$
The value of $d$ at which $y$ attains maxima is given by
$\frac{d y}{d x}=0\,\,\, $(by calculus)
$\Rightarrow -16+34 d-15 d^{2}=0$
$\Rightarrow d=\frac{34 \pm 14}{30}=\frac{2}{3}, \frac{8}{5}$
Now, $\frac{d^{2} y}{d x^{2}}=34-30 d>0 \text { for } d=\frac{2}{3}$
$< 0 $ for $d=\frac{8}{5}$
Hence, $y$ is maximum for $d=\frac{8}{5}$