The sum of n terms of two arithmetic progressions are in the ratio (2n + 3):(7n + 5). Find the ratio of their 9^th terms.(a) 4:5(b) 5:4(c) 9:31(d) 31:9I had been asked this question in final exam.The question is from Arithmetic Progression(A.P.) in section Sequences and Series of Mathematics – Class 11

The sum of n terms of two arithmetic progressions are in the ratio (2n

1.	The sum of n terms of two arithmetic progressions are in the ratio (2n + 3):(7n + 5). Find the ratio of their 9^th terms.(a) 4:5(b) 5:4(c) 9:31(d) 31:9I had been asked this question in final exam.The question is from Arithmetic Progression(A.P.) in section Sequences and Series of Mathematics – Class 11
Answer» The CORRECT choice is (c) 9:31 To explain: Let a, a’ be the first terms and d, d’ be the COMMON differences of 2 A.P.’s respectively. Given, \(\FRAC{\frac{n}{2}[2a+(n-1)d]}{\frac{n}{2}[2a’+(n-1)d’]} = \frac{2n+3}{7n+5}\) =>\(\frac{a+(n-1)d/2}{a’+(n-1) d’/2} = \frac{2n+3}{7n+5}\) If we have to find ratio of 9^th terms then (n-1)/2 =8 => n=17 =>\(\frac{a+8d}{a’+8d’} = \frac{217+3}{717+5} = \frac{34+3}{119+5} = \frac{36}{124}\) = 9/31.

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