Question

The value of the common difference of an A.P. which makes $T_1\ast T_2\ast T_7$ least, given that $T_7=9$ is

• A. $d=\frac{33}{2}$
• B. $d=\frac{5}{4}$
• C. $d=\frac{33}{20}$
• D. none of these

Answer 1

The correct option is C $d=\frac{33}{20}$

Let $D$ be common difference & $a$ be first term

$a+6D=9$

$\Rightarrow$ $a=9-6D$

$P=a.(a+D)(a+6D)$

$P=(9-6D)(9-5D)\left(9\right)$

$\frac{dP}{dD}=0$

$\Rightarrow \frac{d}{dD}\left[(81-99D+30D^2).9\right]=0$

$\Rightarrow 60D-99=0$

$\Rightarrow D=\frac{99}{60}=\frac{33}{20}$ [C]