Let x_n, y_n, z_n, w_n denote the n^{th} term | vedclass

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(A) Let $S_n = x_n + y_n + z_n + w_n$. Since the sum of arithmetic progressions is also an arithmetic progression, let $S_n = A + (n-1)D$.Given $S_4 =

Vedclass · Accepted Answer

$10^4$

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(A) Let $S_n = x_n + y_n + z_n + w_n$. Since the sum of arithmetic progressions is also an arithmetic progression, let $S_n = A + (n-1)D$.Given $S_4 = A + 3D = 8$ and $S_{10} = A + 9D = 20$.Subtracting the first equation from the second: $(A + 9D) - (A + 3D) = 20 - 8 \Rightarrow 6D = 12 \Rightarrow D = 2$.Substituting $D = 2$ into $A + 3D = 8$, we get $A + 6 = 8 \Rightarrow A = 2$.Now, $S_{20} = A + 19D = 2 + 19(2) = 2 + 38 = 40$.By the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality, for positive terms $x_{20}, y_{20}, z_{20}, w_{20}$:$\frac{x_{20} + y_{20} + z_{20} + w_{20}}{4} \geq (x_{20} \cdot y_{20} \cdot z_{20} \cdot w_{20})^{1/4}$.$\frac{40}{4} \geq (x_{20} \cdot y_{20} \cdot z_{20} \cdot w_{20})^{1/4} \Rightarrow 10 \geq (x_{20} \cdot y_{20} \cdot z_{20} \cdot w_{20})^{1/4}$.Raising both sides to the power of $4$, we get $10^4 \geq x_{20} \cdot y_{20} \cdot z_{20} \cdot w_{20}$.Thus, the maximum value is $10^4$.

Let $x_n, y_n, z_n, w_n$ denote the $n^{th}$ 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:

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