If the ratio of the sum of the first n terms of two Aps is (7n + 1) : (4n + 27) then find the ratio of their 9th terms.

Explanation

Let `a_(1) "and" a_(2)` be the first terms and `d_(1) "and" d_(2)` be the common difference of the two Aps respectively.
Let `S_(n) "and"S'_(n)` be the sum of the first n terms of the two Aps and `T_(n) "and" T'_(n)` be their nth terms respectively.
Then, `(S_(n))/(S'_(n)) = (7n + 1)/(4n + 27) rArr ((n)/(2)[2a_(1) + (n-1)d_(1)])/((n)/(2)[2a_(2) + (n-1)d_(2)]) = (7n +1)/(4n +27)`
`(2a_(1) + (n-1)d_(1))/(2a_(2) + (n-1)d_(2)) = (7n + 1)/(4n +27). " "... (i)`
To find the ratio of mth terms, we replace n by (2m-1) in the above expression.
Replacing n by `(2 xx 9-1)`, i.e., 17 on both sides in (i), we get
`(2a_(1) + (17-1)d_(1))/(2a_(2) + (17-1)d_(2)) = (7 xx 17 + 1)/(4 xx 17 +27) rArr (2a_(1) +16d_(1))/(2a_(2) + 16d_(2)) = (120)/(95)`
`rArr (a_(1) + 8d_(1))/(a_(2) + 8d_(2)) = (24)/(19)`
`rArr (a_(1) + (9-1)d_(1))/(a_(2) + (9-1)d_(2)) = (24)/(19)`
`rArr (T_(n))/(T'_(n)) = (24)/(19)`
`therefore ` required ratio = 24 : 19.