Problem:
Let $a_i , i \in \lbrace {1,2,3,4,5} \rbrace$ represent $5$ terms in A.P. with common difference greater than $2$. If $\sum_{i=1}^5 a_i=30$ and $\Pi_{i=1}^5 a_i=3840$, find $a_1\cdot a_5$
I took the most obvious route-writing all $a_i s$ in terms of $d$ and headed to solve the bi-quadratic equation. $d$ gets us the answer.
However, I think there is a more elegant way of solving this(also supported by the fact that the question doesn't want us to evaluate $d$ directly). I couldn't find one. Can you help?
Let $a_1a_5=x$. Then
$$5 (\frac{a_1+a_5}{2})=30 \implies a_1+a_5=12$$ $$a_3=\frac{a_1+a_5}{2}=6$$ $$a_2a_4=(\frac{a_1+a_3}{2})(\frac{a_3+a_5}{2})=\frac14(a_1+6)(6+a_5)=\frac14 (a_1a_5+6[a_1+a_5]+36)=\frac14(x+108)$$
$$a_1a_2a_3a_4a_5=(a_3)(a_1a_5)(a_2a_4)=\frac64 x(x+108)=3840$$
$$x^2+108x-2560=(x+128)(x-20)=0$$ $$x=+20, -128$$
Since $a_1+a_5=2a_1+4d=12$ and $d\gt 2$ then $a_1 \lt 0$ and $a_5 \gt 0$ so $a_1a_5 \lt 0$. Therefore $$a_1a_5=x=-128$$