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(A) The resistance of a wire is given by $R = \rho \frac{L}{A}$.Since all wires have the same dimensions ($L$ and $A$ are constant),the resistance $R$

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$\frac{n(n+1)}{2}$

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(A) The resistance of a wire is given by $R = \rho \frac{L}{A}$.Since all wires have the same dimensions ($L$ and $A$ are constant),the resistance $R$ is directly proportional to the resistivity $\rho$.When $n$ wires are connected in series,the total resistance is $R_{eq} = R_1 + R_2 + . . . + R_n$.Substituting $R = \rho \frac{L}{A}$,we get $\rho_{eq} \frac{L}{A} = \rho_1 \frac{L}{A} + \rho_2 \frac{L}{A} + . . . + \rho_n \frac{L}{A}$.Canceling $\frac{L}{A}$ from both sides,we get $\rho_{eq} = \rho_1 + \rho_2 + . . . + \rho_n$.Given $\rho_1 = 1, \rho_2 = 2, . . . , \rho_n = n$,the equivalent resistivity is $\rho_{eq} = 1 + 2 + 3 + . . . + n$.Using the sum formula for the first $n$ natural numbers,$\rho_{eq} = \frac{n(n+1)}{2}$.

$n$ conducting wires of same dimensions but having resistivities $1, 2, 3, . . . , n$ are connected in series. The equivalent resistivity of the combination is

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