The product of the third by the sixth term of an arithmetic progression (A.P) is $406$. The division of the ninth term of the progression by the fourth term gives a quotient $2$ and a remainder $6$. Find the first term and the difference of the progression.
Let the first term of A.P $= t_1 = a$ and the common difference $= d$
$n^{th}$ term of A.P $= t_n = a + \left(n - 1\right) d$
$\therefore \;$ $3^{rd}$ term of A.P $= t_3 = a + 2d$;
$4^{th}$ term of A.P $= t_4 = a + 3d$;
$6^{th}$ term of A.P $= t_6 = a + 5d$;
$9^{th}$ term of A.P $= t_9 = a + 8d$;
Given: $\;$ $t_3 \times t_6 = 406$
i.e. $\;$ $\left(a + 2d\right) \left(a + 5d\right) = 406$
i.e. $\;$ $a^2 + 7ad + 10d^2 = 406$ $\;\;\; \cdots \; (1)$
Also: $\;$ $t_9 \div t_4$ gives a quotient $2$ and a remainder $6$
i.e. $\;$ $\dfrac{t_9}{t_4} = 2 + \dfrac{6}{t_4}$
i.e. $\;$ $t_9 = 2 t_4 + 6$
i.e. $\;$ $a + 8d = 2 \times \left(a + 3d\right) + 6$
i.e. $\;$ $a + 8d = 2a + 6d + 6$
i.e. $\;$ $a = 2d - 6$ $\;\;\; \cdots \; (2)$
In view of equation $(2)$, equation $(1)$ becomes
$\left(2d - 6\right)^2 + 7 \times \left(2d - 6\right) \times d + 10d^2 = 406$
i.e. $\;$ $4d^2 - 24d + 36 + 14d^2 - 42d + 10d^2 = 406$
i.e. $\;$ $28 d^2 - 66d - 370 = 0$
i.e. $\;$ $14 d^2 - 33 d - 185 = 0$
i.e. $\;$ $d = \dfrac{33 \pm \sqrt{33^2 - 4 \times 14 \times \left(-185\right)}}{2 \times 14}$
i.e. $\;$ $d = \dfrac{33 \pm \sqrt{1089 + 10360}}{28} = \dfrac{33 \pm \sqrt{11449}}{28}$
i.e. $\;$ $d = \dfrac{33}{28} \pm \dfrac{107}{28}$
i.e. $\;$ $d = 5$ $\;$ or $\;$ $d = \dfrac{-37}{14}$
When $\;$ $d = 5$, $\;$ we have from equation $(2)$, $\;$ $a = \left(2 \times 5\right) - 6 = 4$
When $\;$ $d = \dfrac{-37}{14}$, $\;$ we have from equation $(2)$, $\;$ $a = \left[2 \times \left(\dfrac{-37}{14}\right)\right] - 6 = \dfrac{-79}{7}$