This is the assignment. We have two equations which we should use to get the first member of the arithmetical progression and the step difference d,
$$\frac{a_{27}}{a_{5}}=5,\qquad a_{3}\cdot a_{6}=364.$$
I always get a quadratic equation with the solutions $a_{1}=\pm3\sqrt2$ and $d=\pm2\sqrt2$. However, my sister's workbook gives the solutions as $a_1=\pm6$ and $d=\pm4$ . Which is my solution multiplied by $\sqrt2$. I think I made a mistake somewhere but I just can't find it. Please help if you can.
We can start by writing everything in terms of the first term, $a_1$. If $$\frac{a_{27}}{a_5}= 5 \rightarrow a_{27}=5a_5$$ $$a_1+26d=5(a_1+4d)$$ $$a_1 +26d = 5a_1 + 20d$$ $$4a_1 = 6d \rightarrow a_1 = 6d/4$$ Now we write the second condition in terms of $a_1$ and $d$. $$a_3 \cdot a_6 = 364$$ $$(a_1 + 2d)(a_1 + 5d) = 364$$ $$a_1^2 + 7da_1 +10d^2 = 364$$
Now plug in $a_1 = 6d/4$: $$(\frac{6d}{4})^2 + 7d(\frac{6d}{4}) + 10d^2 = 364$$ $$\frac{36d^2}{16} + \frac{42d^2}{4} + 10d^2 = 364$$ $$\frac{36d^2}{16} + \frac{168d^2}{16} + \frac{160d^2}{16} = 364$$ $$364d^2 = 364 \cdot 16$$ $$d^2 = 16 \rightarrow d = \pm 4$$
I hope this helps you identify where you made a mistake. Can you take it from here?