The sum of the first ten terms of an AP is four times the sum of the first five terms, the ratio of the first term to the common difference is

Explanation

Let `P(n):tan^(-1)((1)/(3))+tan^(-1)((1)/(7))+…..+tan^(-1)((1)/(n^+n+1))=tan^(-1)((n)/(n+2))` ……..(i)
Step I For `n=1`
LHS of Eq. (i) `=tan^(-1)((1)/(3)) =tan^(-1)((1)/(1+2))=RHS of Eq. (i)
Therefore , P(1) is true .
Step II Assume that P(k) is true. Then ,
`P(k):tan^(-1)((1)/(3))+tan^(-1)((1)/(3))+tan^(-1)((1)/(7))+........+tan^(-1)((1)/(k^2+k+1))=tan^(-1)((k)/(k+2))`
Step III For `n=k+1`
`P(k+1):tan^(-1)((1)/(3))+tan^(-1)((1)/(7))+....+tan^(-1)((1)/(k^2+k+1))+tan^(-1)((1)/((k+1)^2+(k+1)+1))=tan^(-1)((k+1)/(k+3))`......(ii)
LHS of Eq. (ii)
`=tan^(-1)((1)/(3))+tan^(-1)((1)/(7))+.....+tan^(-1)((1)/(k^2+k+1))+tan^(-1)((1)/((k+1)^2+(k+1)+1))`
`=tan^(-1)((k+1)/(k+2))+tan^(-1)((1)/((k+1)^2+(k+1)+1))` [by aaumption atep]
`=tan^(-1)((k)/(1+(k+1)))+tan^(-1)((1)/(k^2+3k+3))`
`=tan^(-1)((k)/(1+(k+1)))+tan^(-1)((1)/(1+(k+1)(k+2)))`
`=tan^(-1)(((k+1)-1)/(1+(k+1).1))+tan^(-1)(((k+2)-(k+1))/(1+(k+2)(k+1)))`
`=tan^(-1)(k+1)-tan^(-1)1+tan^(-1)(k+2)-tan^(-1)(k+1)=tan^(-1)(k+2)-tan^(-1)1`
`=tan^(-1)((k+2-1)/(1+(k+2).1))=tan^(-1)((k+1)/(k+3))=` RHS of Eq. (ii)
This shows that the result is true for `n=k+1`. Hence. by the principle of mathematical induction . the result is true for all `n in N`.