Determine the first element and the difference of arithmetical sequence

Question

Determine the first element and the difference of arithmetical sequence where the following is: $$a_3+a_{10}=12$$ $$a_6a_{7}=18$$

I tried to solve it this way: $$(a_1+5d)(a_1+6d)=18$$ $$(a_1+2d)+(a_1+9d)=12$$

The thing is that when I multiply the first expression:

$$a_1^2+11da_1+30d^2=18$$ This is where things get complicated...

Answer 1

Your second equation is $$2a_1+11d=12$$ or $$a_1=6-\frac{11d}2.$$ Substituting in your first equation $$(a_1+5d)(a_1+6d)=18$$ gives a quadratic equation for $d$.

Over to you!

Answer 2

Alt. hint: $\;a_6+a_7 = a_3+a_{10}\,$ because the sequence is an AP, and therefore:

$$ \begin{align} a_6 + a_7 = 12 \\ a_6 \cdot a_7 = 18 \end{align} $$

It follows that $\,a_6, a_7\,$ are the roots of the quadratic equation $\,x^2 - 12 x + 18 = 0\,$.