Series + Number theory problem from JEE exam .

Question

My approach : Used variables for given conditions and made some equations. I got the AP as 3,5,7 and GP as 1,3,27 but it is not matching with the condition given in the question. I would really appreciate your help. Thank you.

Answer 1

We have that $$a_2=\frac{a_1+a_3}{2}=\frac{15-a_2}{2}\implies a_2=5, a_1+a_3=10$$ but $a_1,a_3$ are not supposed to be integers. Similarly, $$b_2^2=b_1b_3=\frac{27}{b_2}\implies b_2=3, b_1b_3=9$$ but $b_1,b_3$ are not supposed to be integers. Moreover $$(a_1+b_1)(a_3+b_3)=(a_2+b_2)^2=(5+3)^2=64=2^6$$ Therefore $a_1+b_1=2^{6-n}\leq a_3+b_3=2^n$ with $n=3,4,5,6$. By checking the four cases, it turns out that the good one is $n=6$ and that the maximum value is $$a_3=\frac{73+7\sqrt{73-4\cdot 3}}{2}\approx 63.836.$$ Hence the given answers are both confirmed.