Question

Let $x_n,y_n,z_n,w_n$ denote $n^{\text{th}}$ term of four different arithmetic progressions with positive terms. If $x_4+y_4+z_4+w_4=8$ and $x_{10}+y_{10}+z_{10}+w_{10}=20,$ then maximum possible value of $x_{20}\cdot y_{20}\cdot z_{20}\cdot w_{20}$ is

• A. ${10}^4$
• B. ${10}^8$
• C. ${10}^{10}$
• D. ${10}^{20}$

Answer 1

The correct option is A ${10}^4$

Let $x_n=a_x+(n-1)d_x$
$y_n=a_y+(n-1)d_y$
$z_n=a_z+(n-1)d_z$
$w_n=a_w+(n-1)d_w$

Given $x_4+y_4+z_4+w_4=8$
$\Rightarrow a_x+a_y+a_z+a_w+3(d_x+d_y+d_z+d_w)=8\cdots \left(1\right)$

Also $x_{10}+y_{10}+z_{10}+w_{10}=20$
$\Rightarrow a_x+a_y+a_z+a_w+9(d_x+d_y+d_z+d_w)=20\cdots \left(2\right)$

Solving equation $\left(1\right)$ and $\left(2\right)$. we get
$d_x+d_y+d_z+d_w=2a_x+a_y+a_z+a_w=2$

As the terms of the A.P. is positive, using $A.M\ge G.M$, we get
$\frac{x_{20}+y_{20}+z_{20}+w_{20}}{4}\ge (x_{20}\cdot y_{20}\cdot z_{20}\cdot w_{20}{)}^{1/4}\Rightarrow \frac{a_x+a_y+a_z+a_w+19(d_x+d_y+d_z+d_w)}{4}\ge (x_{20}\cdot y_{20}\cdot z_{20}\cdot w_{20}{)}^{1/4}\Rightarrow 10\ge (x_{20}\cdot y_{20}\cdot z_{20}\cdot w_{20}{)}^{1/4}\Rightarrow x_{20}\cdot y_{20}\cdot z_{20}\cdot w_{20}\le {10}^4$

Hence, the maximum possible value of the given expression is ${10}^4$.