Question

Given that $a_1,a_2,a_3$ is an A.P in that order, $a_1+a_2+a_3=15;b_1,b_2,b_3$ is a G.P. in that order, and $b_1{b}_2{b}_3=27$. If $a_1+b_1,a_2+b_2,a_3+b_3$ are positive integers and form a $G.P$. in that order, then

• A. maximum possible value of $a_3$ is $15(a_1,a_2,a_3$ are positive integer and unequal)
• B. minimum possible value of $a_3$ is $3(a_1,a_2,a_3$ are positive integer and unequal)
• C. maximum possible value of $a_3$ is $9+\frac{55+7\sqrt{61}}{2}$
• D. maximum possible value of $a_3$ is $4+\frac{55+7\sqrt{61}}{2}$

Answer 1

Answer: (A), (B), (C)

Let $a_1=5-d,a_2=5,a_3=5+d,b_1=\frac{3}{q}$, $b_2=3,b_3=3q$,
then the given conditions indicate that $5-d+\frac{3}{q}$ and $5+d+3q$ are all positive integers and
$\left(5-d+\frac{3}{q}\right)(5+d+3q)=64$
It is easy to check that for getting maximum positive d, there are only four
possibilities for $\left(5-d+\frac{3}{q},5+d+3q\right):(1,64),(2,32),(4,16)$ and $(8,8)$.
(i) By solving the system of equations corresponding to $(1,64)$, it follows that
$3q+\frac{3}{q}=55\Rightarrow q_{1,2}=\frac{55\pm 7\sqrt{61}}{6},d_{1,2}=4+\frac{3}{q_{1,2}}$
$\Rightarrow d_{\max }=4+\frac{3.6}{55-7\sqrt{61}}=4+\frac{55+7\sqrt{61}}{2}$
(ii) By solving the system of equations corresponding to $(2,32)$, it follows that
$q+\frac{1}{q}=8\Rightarrow q_{1,2}=\frac{8\pm \sqrt{60}}{2}=4\pm \sqrt{15},d_{1,2}=3+\frac{3}{q_{1,2}}$
$\Rightarrow d_{\max }=3+\frac{3}{4-\sqrt{15}}=15+3\sqrt{15}$
(iii) By solving the system of equation corresponding to $(4,16)$, it follows that $3q+\frac{3}{q}=10\Rightarrow q=3$ or $\frac{1}{3},d_{1,2}=1+\frac{3}{q_{1,2}}=2$ or $10\Rightarrow d_{\max }=10$
(iv) By solving the system of equations corresponding to $(8,8)$, it follows that $q+\frac{1}{q}=2\Rightarrow q=1,d=0$
In summary, $d_{\max }=4+\frac{55+7\sqrt{61}}{2}$, hence the maximum possible value of $a_3$ is $9+\frac{55+7\sqrt{61}}{2}$