Question

Let $x_n,y_n,z_n,w_n$ denote $n^{th}$ terms 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 the maximum value of $x_{20}\cdot y_{20}\cdot z_{20}\cdot w_{20}$ is

• A. ${10}^4$
• B. ${10}^6$
• C. ${10}^8$
• D. ${10}^{10}$

Answer 1

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

We know that sum of two A.P's will also be an A.P, so
$x_4+y_4+z_4+w_4=A+3D=\mathrm{8......}\left(i\right)$
$x_{10}+y_{10}+z_{10}+w_{10}=A+9D=\mathrm{20.....}\left(ii\right)$
Using equation (i) and (ii),
$A=2,D=2$
Now,
$x_{20}+y_{20}+z_{20}+w_{20}=A+19D=40$
We know that $A.M\ge G.M$
$\therefore \frac{x_{20}+y_{20}+z_{20}+w_{20}}{4}\ge (x_{20}\cdot y_{20}\cdot z_{20}\cdot w_{20}{)}^{\frac{1}{4}}$
Hence $x_{20}\cdot y_{20}\cdot z_{20}\cdot w_{20}\le {10}^4$