The sums of n terms of two arithmetic progresssions are in the ratio `(7n+1):(4n+17).` Find the ratio of their nth terms and also common differences.

Explanation

Given, `S_(n):S'_(n)=(7n+1):(4n+17)`
Here, `A=7,B=1,C=4 and D = 17 `
`therefore (t_(n))/(t'_n)=(7(2n-1)+1)/(4(2n-1)+17)=(14n-6)/(8n+13)`
and `(d)/(d')=(A)/(C)=(7)/(4)`
Hence, `t_(n):t'_(n)=(14n-6):(8n+13)` and `d:d'= 7:4`