Question

If $t_n$ be the $n^{th}$ term of an A.P and if $t_7=9$, then the value of the common difference that would make $t_1{t}_2{t}_7$ least, is

• A. $\frac{33}{40}$
• B. $\frac{33}{20}$
• C. $\frac{33}{10}$
• D. none of these

Answer 1

The correct option is B $\frac{33}{20}$

Let the first term be $a$, and the common difference $x$.
Hence $t_1=a$ , $t_2=a+2x$
Now $t_7=a+6x=9$
Hence
$a=9-6x$
Therefore $t_1.t_2.t_7=\alpha =\left(a\right)(a+x)\left(9\right)=(9-6x)(9-5x)\left(9\right)$
Or
$\alpha =9[81-99x+30x^2]$
Now
$\frac{d\alpha }{dx}=60x-99=0$
$x=\frac{33}{20}$.
$\frac{d^2\alpha }{d{x}^2}=60$
Now $\alpha ">0$
Hence minima.
Therefore $t_1,t_2,t_7$ is least at
$x=\frac{33}{20}$.