Question

The value of the common difference of an arithmetic progression, which makes $T_1{T}_2{T}_7$ the least. Given that, $T_7=9$ is

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

Answer 1

The correct option is C

$\frac{33}{20}$

Explanation for the correct option

Let us assume that, the common difference is $D$.

It is given that, $T_7=9$.

Thus, $T_7-T_1=\left(7-1\right)D$

$$\begin{gathered} \Rightarrow T_1=T_7-6D \\ \Rightarrow T_1=9-6D \end{gathered}$$

Therefore, $T_7-T_2=\left(7-2\right)D$

$$\begin{gathered} \Rightarrow T_2=T_7-5D \\ \Rightarrow T_2=9-5D \end{gathered}$$

So, $T_1{T}_2{T}_7=\left(9-6D\right)\left(9-5D\right)\left(9\right)$.

$$\begin{gathered} \Rightarrow T_1{T}_2{T}_7=9\left(81-99D+30D^2\right) \\ \Rightarrow T_1{T}_2{T}_7=270D^2-891D+729 \end{gathered}$$

Let us assume that, $T_1{T}_2{T}_7=f\left(D\right)$

$\Rightarrow f\left(D\right)=270D^2-891D+729$

Differentiate both sides of the equation with respect to $d$.

$$\begin{gathered} \Rightarrow \frac{d}{dD}\left[f\left(D\right)\right]=\frac{d}{dD}\left(270D^2-891D+729\right) \\ \Rightarrow f\text{'}\left(D\right)=540D-891 \end{gathered}$$

For the least value of $f\left(D\right)$, $f\text{'}\left(D\right)=0$.

$$\begin{gathered} \Rightarrow 540D-891=0 \\ \Rightarrow D=\frac{891}{540} \\ \Rightarrow D=\frac{33}{20} \end{gathered}$$

Hence, option C is correct.