\(K_{eq}=\frac{ K _{1} A _{1}+ K _{2} A _{2}}{ A _{1}+ A _{2}}\)

Cross sectional Area \(A _{1}= A _{2}= A\), for all rods.

for any two rods having same coefficient, \(K _{1}\) the resultant is also \(K _{1}\)

so the above combination will reduce to a combination having just two rods one with \(K _{1}\) and another with \(K _{2}\)

So net coefficient of conductivity will be \(K=\frac{K_{1}+K_{2}}{2}\)