ratio of arithmetic summation This is the question which I am referring to

Find the AP in which the ratio of the sum to n terms to the sum of succeeding n terms is independent of n.

What I have thought:
we are talking about the following ratio 
$\frac{\frac{n}{2}[2a+(n-1)d]}{\frac{n}{2}[2(a+nd)+(n-1)d]}$
now for this ratio to be independent of $n$ i think $d=0$ but the solution of this answer is : $k,3k,5k... $ for any value of $k$ .I dont know how come this result came .Any help will be appreciated.
 A: You are almost there. Slightly manipulating the ratio you have obtained gives
$$\frac{2a+(n-1)d}{2(a+nd)+(n-1)d}=\frac{2a-d+nd}{2a-d+3nd}$$
For this expression to be independent of $n$, you need to be able to cancel the $n$ in the nominator with the $n$ in the denominator. This happens exactly when $2a-d=0$. 
This condition is equivalent to the desired result, which you can see by setting $d=2k$ (and thus $a=k$, giving the progression $k,3k,5k,\dots$).
A: Let the progression be given by $a_n = a_0 + n d$. Then we have $$\frac {a_1 + a_2 + \cdots + a_n} {a_{n + 1} + a_{n + 2} + \cdots + {a_{2n}}} = c,$$ where $c$ is a constant. This means, $$\frac {2 (a_1 + a_2 + \cdots + a_n)} {a_1 + a_2 + \cdots + a_{2n}} = c.$$ That, in turn, is $$c = \frac {2 n a_0 + n (n + 1) d} {2 n a_0 + n (2n + 1) d} = \frac {1} {2} + \frac {2 a_0 + d} {4 a_0 + (4n + 2) d}.$$ Hence, $d = -2 a_0$. So, we have $a_n = (2 n - 1) d/2$. Put $d = 2k$ to have $a_n = (2n - 1)k$ and $a_0 = -k$. Then the progression is: $-k, k, 3k, 5k, \cdots$.
