If $a_1,a_2,\mathrm{....}{a}_n$  are in A.P. with $a_7=9$  and $a_1{a}_2{a}_7$ is minimum, then common difference of A.P. is:

$a_1+6d=9.$  Let $P=a_1{a}_2{a}_7.$  Then $P=a_1\left(a_1+d\right)\times 9=9a_1\left[a_1+\left(9-a_1\right)/6\right]$

$=\frac{9}{6}\left[5a_1^2+9a_1\right]=\frac{45}{6}\left\{{\left(a_1+\frac{9}{10}\right)}^2-\frac{81}{100}\right\}$

$\therefore$     $P$ is minimum when $a_1=-9/10$