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The sum of n terms of two A.P.’s are in the ratio 5n + 4 : 9n + 6. Then, the ratio of their 18th term is

A. $\frac{179}{321}$ B. $\frac{178}{321}$ C. $\frac{175}{321}$ D. $\frac{176}{321}$ Answer: Option A
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Solution(By Apex Team)

Let a1, d2 be the first terms of two ratios S and S’ and d1, d2 be their common difference respectively Then, $\begin{aligned}&S_n=\frac{n}{2}\left[2a_1+(n-1)d_1\right]\text{ and }\\ &S_n^{\prime}=\frac{n}{2}\left[2a_2+(n-1)d_2\right]\\ &\text{ Now }\\ &\frac{S_n}{S_n^{\prime}}=\frac{\frac{n}{2}\left[2a_1+(n-1)d_1\right]}{\frac{n}{2}\left[2a_2+(n-1)d_2\right]}\\ &=\frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}\\ &\text{ But }\frac{S_n}{S_n^{\prime}}=\frac{5n+4}{9n+6}\\ &\therefore\frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}=\frac{5n+4}{9n+6}\end{aligned}$ Now we have to find the ratios in 18th term Here n = 18 $\begin{aligned}\therefore\frac{2a_1+(18-1)d_1}{2a_2+(18-1)d_2}&=\frac{5(2n-1)+4}{9(2n-1)+6}\\ =&\frac{5(2\times18-1)+4}{9(2\times18-1)+6}\\ =&\frac{5\times35+4}{9\times35+6}\\ =&\frac{175+4}{315+6}\\ =&\frac{179}{321}\end{aligned}$

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