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.
 A: You are almost there
$$
a_{n+1} = a_1 + nd\\
a_n = a_1 + (n-1)d
$$
thus we get 
$$
a_{n+1}a_n = \left(a_1 + nd\right)\left(a_1 + (n-1)d\right)
$$
re-arrange your given relation you find
$$
a_{n+1}a_n =9n^2-21n + 10
$$
can you equate coefficients?
A: We have
$$a_2=\frac{9-21+10}{a_1}\Rightarrow a_1a_2=-2\tag1$$
and
$$a_3=\frac{36-42+10}{a_2}\Rightarrow a_2a_3=4\tag2$$
Since we have $a_1+a_3=2a_2$, with $(1)(2)$, we have
$$a_1+\frac{4}{a_2}=2a_2\Rightarrow a_1a_2+4=2a_2^2\Rightarrow a_2=\pm1.$$
Can you take it from here?
A: $$ a_{n+1}=\frac{9n^2-21n+10}{a_n}\\a_{n+1}a_{n}=9n^2-21n+10\\(a_1+nd)(a_1+(n-1)d)=9n^2-21n+10\\a_1^2+(2n-1)a_1d+(n^2-n)d^2=9n^2-21n+10\\n^2d^2+n(-d^2+2a_1d)+(a_1^2-a_1d)=9n^2-21n+10$$ and now $$\left\{\begin{matrix}
d^2=9 & \\ 
 -d^2+2a_1d=-21& \\ 
 a_1^2-a_1d=10& 
\end{matrix}\right. $$ you can find $d=\pm 3$ first ,and put d to find $a_1$
A: The formula for an arithmetic progression is in the form of of a line $mn+b$. According to your equation $(m(n+1)+b)(mn+b)$ = $9n^2-21n+10$
$(mn+m+b)(mn+b)$=$(3n+3-5)(3n-5)$. Therefore  $a_n$=$3n-5$. However if we consider $(mn+m+b)(mn+b)$=$(-3n-3+5)(-3n+5)$ then $a_n$ = $-3n+5$ . The reason we have two possible answers is the first coiffecent a of both factors need to multiply to $9$. $m^2$=$9$ has solutions of both positive and negative three.
A: 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 $a_3 = a_2$.
