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

Explanation

Let `a_(1),a_(2)` and `d_(1),d_(2)` be the first terms and the common difference of the first and second arthimetic progression respectively.
According to the given condition, we have
`(n/2[2a_(1)+(n-1)d_(1)])/(n/2[2a_(2)+(n-1)d_(2)])=(5n+4)/(9n+6)`
`rArr(2a_(1)+(n-1)d_(1))/(2a_(2)+(n-1)d_(2))=(5n+4)/(9n+6)`
Substituting n=35 in (1), we get
`(2a_(1)+34d_(1))/(2a_(2)+34d_(2))=(5(35)+4)/(9(35)+6)`
or `(a_(1)+17d_(1))/(a_(2)+17d_(2))=179/321`
or `("18th term of first A.P")/("18th term of second A.P")=179/321`
Thus, the ratio of 18th term of both the A.P.'s is 179:321.