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 $15\left(a_1, a_2, a_3\right.$ are positive integer and unequal)
• B. minimum possible value of $a_3$ is $3\left(a_1, a_2, a_3\right.$ 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 $a_1=5- d , a _2=5, a _3=5+ d , b _1=\frac{3}{ q }$, $b _2=3, b _3=3 q$,
then the given conditions indicate that $5- d +\frac{3}{ q }$ and $5+ d +3 q$ are all positive integers and
$ \left(5-d+\frac{3}{q}\right)(5+d+3 q)=64 $
It is easy to check that for getting maximum positive d, there are only four
possibilities for $\left(5- d +\frac{3}{ q }, 5+ d +3 q \right):(1,64),(2,32),(4,16)$ and $(8,8)$.
(i) By solving the system of equations corresponding to $(1,64)$, it follows that
$3 q +\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 $3 q +\frac{3}{ q }=10 \Rightarrow q =3$ 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 $9+\frac{55+7 \sqrt{61}}{2}$