STEP is a Cambridge University devised examination paper for mathematics. I have been solving Q1 from the STEP III paper
And I have stumbled upon a 'weird', not even sure whether it is correct, way of proving that the solutions are in a geometric progression: solution (Scroll down and it is STEP III Q1). Not sure how the solution of that person proves that the roots are in geometric progression. This is my question 1.
My second question concerns what necessary and sufficient means in the final part of the first question.
I have reasoned in the following way, when solving that part. Let $T_1,T_2,T_3$ be the three consecutive terms of arithmetic progression such that $T_1=ak^{-1}$, $T_2=a$, $T_3=ak$. We know that the sum of $n$ first terms in arithmetic progression is $\frac{n(2T_1+n(T_2-T_1))}{2}$, then we have that:
$$ak^{-1}+a+ak=\frac{3}{2}(2ak^{-1}+3(a-ak^{-1}))$$ $$a(1+k^{-1}+k)=\frac{3a(3-k^{-1})}{2}$$ which can be further simplified: $$\frac{a}{2}(2k+5k^{-1}-7)=0$$
Now I am not sure whether this is necessary and/or sufficient, since I do not know what those mean exactly.
For the second question, choose the roots to be $a-k, a, a+k$ (as the number of roots is odd)
For the even number of terms in arithmetic progression, we should choose $a\pm k,a\pm2k,\cdots$
$\implies a-k+a+a+k=p\iff a=\dfrac p3$
and $a(a-k)(a+k)=r\iff27r=p(p-3k)(p+3k)\iff9k^2=\dfrac{p^3-27r}p$
Finally $q=a(a-k)+(a-k)(a+k)+a(a+k)=3a^2-k^2$
Replace the values of $a,k^2$