Question

The product of the third by the sixth term of an arithmetic progression 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.

Answer 1

Let $a$ be the first term and $d$ be the common difference and we know that $n$th term, $T_n=a+(n-1)d$

$\therefore T_3=a+2d\&T_6=a+5d{T}_3\times T_6=406$

$(a+2d)(a+5d)=406$

$a^2+2ad+5ad+10a^2=406$

$a^2+7ad+10d^2=406-\left(i\right)Now,T_9=T_4\times 2+6$

$a+8d=(a+3d)2+6$

$a+8d=2a+6d+6$

$a-2d+6=0$

$=2d-6-\left(ii\right)$

Replacing $a$ from $\left(ii\right)$ in $\left(i\right)$.

${(2d-6)}^2+7(2d-6)d+10d^2=406$

$4d^2-24d+36+{14d}^2-42d+10d^2=406$

${28d}^2-66d+36=406$

$14d^2-33d+18=203$

${14d}^2-33d-185=0$

Using quadratic equation:

$d=\frac{-(-33)\pm \sqrt{{(-33)}^2-4\times 14\times (-185)}}{2\times 14}=\frac{33\pm \sqrt{{\left(33\right)}^2+4\times 14\times 185}}{28}=\frac{33\pm \sqrt{1089+10360}}{28}=\frac{33\pm 107}{28}d=\frac{33+107}{28},\frac{33-107}{28}=\frac{140}{28},\frac{-74}{28}=5,\frac{-37}{28}$

But common difference cannot be negative in this case.

$\therefore$ Common difference$=5$.

From $\left(ii\right)$

$a=2d-6=2\times 5-6=10-6=4$

$\therefore$ The first term is $4$.