Please submit your concern

Explanation:

First series: (S₁) = x, y, $\frac{x}{2},$ z, x + 20

Second series: (S₂) = a₁, a₂, a₃, a₄

Now a₁ = y – x, a₂ = $\frac{x}{2}$ - y, a₃ = z - $\frac{x}{\mathrm{2 }}$and a₄ = x + 20 – z

a₂ – a₁ = 30 gives 3x – 4y = 60 ... (1)

a₄ – a₃ = 30 gives 3x – 4z = 20 ... (2)

and a₄ – a₂ = 60 gives x – 2z + 2y = 80 ... (3)

Solving these equations we get the values of x = 100, y = 60, z = 70

∴ S₁ is 100, 60, 50, 70, 120

S₂ is –40, –10, 20, 50.

Hence, option (c).