The ratio between the sum of first n terms of two A.P.'s are in the ratio (7n-5) : (5n+17). Find the ratio of their 10th terms.

Explanation

Let `a_(1)` and `d_(1)` be the first term and common difference of first A.P. and let `a_(2)` and `d_(2)` be the first term and common difference of other A.P.
`:. ((n)/(2)[2a_(1)+(n-1)d_(1)])/((n)/(2)[2a_(2)+(n-1)d_(2)])=(7n-5)/(5n+17)`
`rArr (2a_(1)+(n-1)d_(1))/(2a_(2)+(n-1)d_(2))=(7n-5)/(5n+17)`
`rArr (a_(1)+((n-1)/(2))d_(1))/(a_(2)((n-1)/(2))d_(2))=(7n-5)/(5n+17) " " ...(1)` (dividing Nr. and Dr. by 2)
Replace `(n-1)/(2)` by 9 `" "` (`:' (T_(10))/(T_(10)^(**))=(a_(1)+9d_(1))/(a_(2)+9d_(2))`)
`rArr (n-1)/(2)=9 rArr n=19`
`:. ` From eq. (1),
`(a_(1)+9d_(1))/(a_(2)+9d_(2))=(7(19)-5)/(5(19)+17)`
`rArr (T_(10))/(T_(10)^(**))=(128)/(112)=(8)/(7)`
`:. T_(10):T_(10)^(**)= 8:7`.