2 wires of same dimension but different resistivity(rho 1 & rho 2) are connected in series and after that in parallel,then the equivalent resistivity of the conductor in both the case will be?

series: rho(series)= (rhoi+rho2)/2

parallel:rho(parallel)=(rho1)(rho2)/(rho1 + rho2)

R1=(rho1)L/A

R2=(rho2)L/A

In series; R_eq=(rho1+rho2)L/A =>Resistivity=(rho1+rho2)

In parallel; R_eq=((rho1)(rho2)/(rho1+rho2))L/A =>Resistivity=R_eq(parallel) X (A/L)

my previous solution was wrong. the correct answer is:

series: (rho1 + rho2)/2

parallel: (rho1*rho2)/(2*(rho1+rho2))

please inform me in case you want more explanation

yes sir ,i want to know how to approach and solve this question exactly instead of just knowing the answers.

R₁= ρ₁₍l/A)

R₂= ρ₂(l/A)

R= ρ(2l)/A

and Rs = R1+R2

» ρs = (ρ₁+ρ₂)/2

and R_p = (R₁.R2)/(R₁+R2)

» ρ_p = (1/2)(ρ₁ρ2)/(ρ₁+ρ2)