An A.P. has even number of terms, the first term of which is unity. $S_1$ and $S_2$ stand for the sum of its first $n$ terms and sum of its last $n$ terms such that $S_1$/$S_2$=$\lambda$. What is the common difference of A.P.?
My attempt: Let d be the common difference. $$S_1=n(2+(n-1)d)/2$$
$$S_2=S_{2n}-S_1$$
$$=2n(2+(2n-1)d)/2-n(2+(n-1)d)/2$$.
Putting $S_2$ as $S_1/\lambda$, I get $$d=2(\lambda-1)/(\lambda+n-1-3n\lambda)$$
I thought this would be the answer, but the answer is given as $2$. Don't know how.
Let the AP have $2n$ terms, $S_n$ denote the sum of first $n$ terms, $a$ be the first term and $d$ be the common difference.
$$S_n=\frac{n}{2}(2a+(n-1)d)$$
As the first term is unity, $a=1$
$$S_1=S_n=\frac{n}{2}(2+(n-1)d)\tag1$$
Also, sum of all $2n$ terms is,
$$S_{2n}=\frac{2n}{2}(2+(2n-1)d)\tag2$$
Now, the sum of last $n$ (say $S_N$) terms would be $$S_2=S_N=S_{2n}-S_n$$
$$S_2=S_N=\frac{2n}{2}(2+(2n-1)d)-\frac{n}{2}(2+(n-1)d)$$
$$\lambda=\frac{S_1}{S_2}$$
$$\lambda=\frac{\frac{n}{2}(2+(n-1)d)}{\frac{2n}{2}(2+(2n-1)d)-\frac{n}{2}(2+(n-1)d)}$$
Flipping the fraction for easier calculation;
$$\frac{1}{\lambda}=\frac{\frac{2n}{2}(2+(2n-1)d)-\frac{n}{2}(2+(n-1)d)}{\frac{n}{2}(2+(n-1)d)}$$
Adding 1 on either sides after simplifying;
$$\frac{1+\lambda}{\lambda}=\frac{2+2nd-d}{2+nd-d}$$
Re-arrange the terms to get $d=f(\lambda)$
$$d=\frac{2-2\left(\frac{1+\lambda}{\lambda}\right)}{1-\frac{1+\lambda}{\lambda}+\left(\frac{1+\lambda}{\lambda}-2\right)n}\tag3$$
Using the fact that in any AP, the common difference does not depend on $"n"$ as in d is not a function of $n$, Implying the above expression has to be free of $n$, to achieve this, $$\frac{1+\lambda}{\lambda}=2$$ which is satisfied for $\lambda=1$
Put $\lambda=1$ in (3), you get common difference as 2.