<s>[INST] Arithmetic Progression. Q. The ratio between the sum of $n$ terms of two A.P's is $3n+8:7n+15$. Find the ratio between their $12$th term.
My method:
Given:
$\frac{S_n}{s_n}=\frac{3n+8}{7n+15}$
$\frac{S_n}{3n+8}=\frac{s_n}{7n+15}=k$
$\frac{T_n}{t_n}=\frac{S_n-S_{n-1}}{s_n-s_{n-1}}=\frac{k\left(\left(3n+8\right)-\left(3\left(n-1\right)+8\right)\right)}{k\left(\left(7n+15\right)-\left(7\left(n-1\right)+15\right)\right)}=\frac{3}{7}$
As this applies for any term:
$\frac{T_{12}}{t_{12}}=\frac{3}{7}$
But this is not the answer. The actual answer is $\frac7{16}$. I know how to obtain that answer.
But why is my solution wrong? Its probably the concept I guess. 
 [/INST] Hint.
We have that
$$
S_n=A_1+\cdots+A_n,
$$
where $A_n=Kn+L$, and hence
$$
S_n=K\frac{n(n+1)}{2}+Ln.
$$
Similarly, $s_n=a_1+\cdots+a_n$, where $a_n=kn+\ell$ and $s_n=k\frac{n(n+1)}{2}+\ell n$,
and
$$
\frac{S_n}{s_n}=\frac{K\frac{n(n+1)}{2}+Ln}{k\frac{n(n+1)}{2}+\ell n}=\frac{K(n+1)+2L}{k(n+1)+2\ell}=\frac{Kn+2L+K}{kn+2\ell+k}=\frac{3n+8}{7n+15},
$$
and
$$
\frac{A_{12}}{B_{12}}=\frac{12K+L}{12k+\ell}
$$
</s>