Question

The sum of first 'n' terms of 2 A.P.s are in the ratio (5n+4):(9n+6).Find the ratio of their 18th term.

Answer 1

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Answer 2

The sum of first 'n' terms of 2 A.P.s are in the ratio (5n+4):(9n+6).Find the ratio of their 18th term.

Let the two A.P.s be a, a+d, a+2d,......, a+(n-1) d
And A, A+D, A+2D,.....,A+(n-1)d

Given :---
$\frac{ \frac{n}{2} [ 2a + (n - 1)d ] }{ \frac{n}{2} [ 2 A + (n - 1) D] } = \frac{5n + 4}{9n + 6}$
$therefore \: \frac{2a + (n - 1)d}{2 A + (n - 1) D } = \frac{5n + 4}{9n + 6}$
Now, here we need to apply some thought
18th term is (a+17d) and (A+17D) in both A.P.s
So, we need to put such a value of n, so that we get 18th terms on LHS

Put n=35

$\frac{2a + 34d}{2 A + 34D } = \frac{5(35) + 4}{9(35) + 6}$
$\frac{2(a + 17d)}{2( A + 71 D)} = \frac{179}{321}$
$\frac{a + 17d}{ A + 17 D } = \frac{179}{321}$
=>Ratio of 18th terms is 179:321