Sequence and Series

Concept:Use the ratio of sums to find when the $n$th terms of two APs are equal.Explanation:Let the APs have first terms $a_1, a_2$ and common differences $d_1, d_2$.Sum of $n$ terms: $S_n = \frac{n}{2}[2a + (n-1)d]$.Given $\frac{S_{1n}}{S_{2n}} = \frac{7n-5}{5n+17}$. Cancel $\frac{n}{2}$:$\frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{7n-5}{5n+17}$.Set $x = n-1$. Then $\frac{2a_1 + x d_1}{2a_2 + x d_2} = \frac{7(x+1)-5}{5(x+1)+17} = \frac{7x+2}{5x+22}$.Let $A = 2a_1, B = d_1, C = 2a_2, D = d_2$. So $\frac{A + Bx}{C + Dx} = \frac{7x+2}{5x+22}$.Cross-multiply: $(A+Bx)(5x+22) = (C+Dx)(7x+2)$.Expand: $5Ax + 22A + 5Bx^2 + 22Bx = 7Cx + 2C + 7Dx^2 + 2Dx$.Equate coefficients:$x^2$: $5B = 7D \implies D = \frac{5}{7}B$.$x$: $5A + 22B = 7C + 2D$.Constant: $22A = 2C \implies C = 11A$.Substitute $C$ and $D$ into $x$ coefficient equation: $5A + 22B = 7(11A) + 2(\frac{5}{7}B) = 77A + \frac{10}{7}B$.Simplify: $22B - \frac{10}{7}B = 77A - 5A \implies \frac{144}{7}B = 72A \implies B = \frac{7}{2}A$.Then $D = \frac{5}{7} \cdot \frac{7}{2}A = \frac{5}{2}A$.We need the $n$th terms equal: $a_1 + (n-1)d_1 = a_2 + (n-1)d_2$. Multiply by 2: $2a_1 + 2x d_1 = 2a_2 + 2x d_2$ i.e. $A + 2xB = C + 2xD$.Substitute: $A + 2x \cdot \frac{7}{2}A = 11A + 2x \cdot \frac{5}{2}A \implies A + 7xA = 11A + 5xA$.Cancel $A$ (non-zero): $1 + 7x = 11 + 5x \implies 2x = 10 \implies x = 5$.Thus $n = x+1 = 6$.Answer:The 6th terms of the two series are equal.