Ratio and Proportion

Concept:Arithmetic means inserted between two numbers form an arithmetic progression (AP).Explanation:Let $d$ be the common difference between consecutive terms.The total number of terms in the AP is $(n+2)$: first term $3$, last term $31$.So, $3 + (n+1)d = 31$, giving $d = \frac{28}{n+1}$.The $k$-th mean is $3 + kd$.Thus, $3^{rd}$ mean $= 3 + 3d$ and $(n-1)^{th}$ mean $= 3 + (n-1)d$.Given the ratio $\frac{3+3d}{3+(n-1)d} = \frac{1}{3}$.Cross-multiplying: $3 + (n-1)d = 3(3+3d) = 9 + 9d$.Simplify: $3 + (n-1)d - 9 - 9d = 0 \implies (n-10)d = 6$.Substitute $d = \frac{28}{n+1}$: $\frac{28}{n+1}(n-10) = 6$.Multiply both sides by $(n+1)$: $28(n-10) = 6(n+1)$.Expand: $28n - 280 = 6n + 6$.Bring terms: $28n - 6n = 6 + 280 \implies 22n = 286$.Hence, $n = \frac{286}{22} = 13$.Answer:$n = 13$, which corresponds to option C.