Question

The sums of n terms of two arithmetic progressions are in the ratio `5n + 4 : 9n + 6`. Find the ratio of their `18^(t h)`terms.

Answer 1

Let A and B are those two A.P. with a and b their respective starting term and `d_a` and `d_b` with their respective difference between each term.
Then,
`(n/2(2a+(n-1)d_a))/(n/2(2b+(n-1)d_b)) = (5n+4)/(9n+6)`
`(2a+(n-1)d_a)/(2b+(n-1)d_b) = (5n+4)/(9n+6)`
`(2a-d_a+nd_a)/(2b-d_b+nd_b) = ((5n+4)k)/((9n+6)k)`
Here, `k` is common factor.
Now, comparing numerator and denominator
`d_a = 5k and 2a-d_a = 4k` and
`d_b= 9k and 2b-d_b= 6k`
`d_a = 5k and a = (9k)/2` and
`d_b= 9k and b= (15k)/2`
Ratio of their `18^(th)` term =
`(((9k)/2)+17(5k))/(((15k)/2)+17(9k)) = 94/321`