Question

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) 31:9

I 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 1

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

=>\(\frac{a+8d}{a’+8d’} = \frac{2*17+3}{7*17+5} = \frac{34+3}{119+5} = \frac{36}{124}\) = 9/31.