sequences...find it

Let the first term of the A.P. = a

and the common difference = d

so T(7)= a + 6d = 9

=> $a= 9 - 6d$

$T(1) \times T(2) \times T(7)= a (a + d).9$ ...........................(1)

Putting $a= 9 - 6d$ in the equation $(1)$

$T(1) \times T(2) \times T(7)= (9 - 6d) (9 - 6d +d ) .9$

Now $T(1) \times T(2) \times T(7)$ is a function of d ..

.so let $T(1) \times T(2) \times T(7)=f(d)$ [ f(d) means a function of d ]

so $f(d) = 9.(9-6d)(9-5d)$

So for getting the least value of f(d), lets differentiate it..and then equate to zero , from there we will get the value of d at which f(d) will be minimum.

$f'(d) = 9(9-6d)(-5) + 9(9-5d)(-6)$

putting $f'(d)=0$ , we get $d=\frac{33}{20}$

and also $f''(d) > 0$ , so at $d=\frac{33}{20}$ . f(d) will attain minimum value.

Hence $\boxed{d=\frac{33}{20}}$