Where i'm doing wrong in the solution? I'm learning Progression and Series with myself.At current about Arithmetic Progression.I'm stuck in a problem from sometime.I don't know why my answer is coming different from book.

Problem-



  The sum of n terms of two A.P's are in the ratio of 7n+1:4n+27 , find the ratio of their 11th term.
  


My solution-
To find the ratio of 11th term i used formula of finding Tn= Sn-Sn-1 .First I calculated ratio of sum of 11th term and 10th term and calculated the difference.
  

Answer in the book :- 4/3
We are given :

$\dfrac{\frac{n}{2}(2a + (n-1)d)}{\frac{n}{2}( 2a' + (n-1)d')} = \dfrac{7n +1}{4n+27}$.

All you have to do is to put $n = 21$ and you will get

$\dfrac{2a + 20d}{2a' + 20d'} = \dfrac{148}{111} = \dfrac{4}{3}$

which is nothing but 

$\dfrac{a + 10d}{a' + 10d'} = \dfrac{a + (11-1)d}{a' + (11-1)d'} = \dfrac{T_{11}}{T'_{11}} = \color{blue}{\dfrac{4}{3}}$
$\frac {T_{11}}{T'_{11}} = \frac{S_{11}-S_{10}}{S'_{11}-S'{10}}\ne \frac{S_{11}}{S'_{11}} - \frac{S_{10}}{S'_{10}} = \frac{7n+1}{4n + 27} - \frac{7(n-1) + 1}{4(n-1) + 27}$

So your answer is way off.

$T_n = a_1 + (n-1)*K; S_n = n*a_1 + \frac{n(n-1)}2K$

$T'_n = b_1 + (n-1)*M; S_n = n*b_1 + \frac{n(n-1)}2M$

So $S_n/S'_n = \frac{n*a_1 + \frac{n(n-1)}2K}{n*b_1 + \frac{n(n-1)}2M} =\frac{a_1 + \frac{(n-1)}2K}{b_1 + \frac{(n-1)}2M}=\frac{2a_1 + (n-1)K}{2b_1 + (n-1)M}=\frac{T_n + a_1}{T'_n + b_1}$

So $\frac{T_{11} + a_1}{T'_{11} + b_1}=\frac{77-1}{44+27}$

But we also have $S_1/S'_1 = a_1/b_1 = \frac{7-1}{4+27}$