finding $a_1$ in an arithmetic progression Given an arithmetic progression such that: $$a_{n+1}=\frac{9n^2-21n+10}{a_n}$$
How can I find the value of $a_1$?
I tried using $a_{n+1}=a_1+nd$ but I think it's a loop..
Thanks.

We are given the condition $a_{n+1}=\dfrac{9n^2-21n+10}{a_n}$ or more compactly
$a_{n} a_{n+1}=9n^2-21n+10 \tag{1}$
Solving for d
By evaluating (1) at $n-1$ we are able to get another term with an $a_n$ factor:
$a_{n-1} a_{n}=9(n-1)^2-21(n-1)+10 = 9n^2-39n+40 \tag{2}$
so by subtracting (1) - (2) : $(a_{n+1} - a_{n-1}) a_{n}=18n-30$ but $a_{n+1} - a_{n-1} = 2d$ so
$2d \cdot a_n = 18n - 30$ so at $n+1$ we have $2d \cdot a_{n+1} = 18(n+1) - 30 = 18n - 12$
$2d \cdot (a_{n+1}-a_n) = 18$ or $2d^2 = 18$ so $\boxed{d = \pm 3}$
For d=3
Now by (1), with $a_n = a_1 + 3n - 3$ and $a_{n+1} = a_1 + 3n$ we have
$(a_1+3n-3)(a_1+3n) = 9n^2-21n+10$
which (eventually) reduces to
${a_1}^2 + 6na_1 - 3a_1 = -12n+10$
Since this must be true for all $n$, we equate coefficients of $n$ (and check the rest) to see the only solution is $\boxed{a_1=-2}$
For d=-3
Now by (1), with $a_n = a_1 - 3n + 3$ and $a_{n+1} = a_1 - 3n$ we have
$(a_1-3n+3)(a_1-3n) = 9n^2-21n+10$
which reduces to
${a_1}^2 - 6na_1 + 3a_1 = -12n+10$
Since this must be true for all $n$, we equate coefficients of $n$ (and check the rest) to see the only solution is $\boxed{a_1=2}$
Remarks
If when solving for $a_1$ (for a given value of $d$) you ony apply (1) to a single $n$, say $n=1$, you will get two possible values for each value of $d$, and hence four solutions for $a_1$. Two of these are incorrect and will give $a_3 = a_2$.
