An increasing arithmetic sequence of positive integers has $ {a_{19}}=20$ and ${a_{a_{20}}}=22$ . Find ${a_{2019}}$ Question:- An increasing arithmetic sequence of positive integers has $ {a_{19}}=20$ and ${a_{a_{20}}}=22$ . Find ${a_{2019}}$
I Found this question on a blogspot.On Seeing the statement of question it looks difficult to solve since we have only small information to proceed.I don't know whether it is possible to find ${a_{2019}}$.
This questiom was asked QSHS PMO Team Selection Test, 2019.
Can anybody help me out!
 A: If $d$ is the common difference of successive terms, then $d$ must be a positive integer; since $a_{19}=20$, this means that $a_{20}\ge 21$, $a_{21}\ge 22$, and $a_n\ge 23$ for all $n>21$. You know that $a_{20}=a_{19}+d=20+d$, so $a_{a_{20}}=a_{20+d}=22$, and therefore $20+d\le 21$. It follows that $d=1$, and since $a_{19}=20$, this implies that $a_n=n+1$ for all $n\ge 1$ and hence that $a_{2019}=2020$.
A: It's arithmetic:
So there is a $d$ and an $a_0$ so that $a_k = a_0+ k*d$.
So we have $a_{19} = a_0 + 19d = 20$.
And $a_{a_{20}} = a_{a_0 + 20d} = a_0 + (a_0 + 20d)*d = 22$
So you are asked what is $a_{2019} = a_0+2019d$.
Which... just do it.  Solve for $d$ solve for $a_0$.

 $a_0 = 20-19d$ so $(20-19d) + (20-19d + 20d)*d = 20-19d + (20+d)d = d^2+d + 20=22$ so $d^2 + d - 2=0$ so $(d-1)(d+2) = 0$ and $d = -2$ of $d = 1$.  But as the sequence is increasing $d=1$ and $a_0 = 20 -19= 1$.

If you want to make it a tiny bit easier note that $a_{20} = a_{19} + d$ so $a_{a_{20}} = a_{19+d} = a_0 + (19+d)*d = 22$.
Also given that $a_k$ is increasing and of integers we know $d$ must be a positive integer but for $a_{19}$ do only be as high as $20$; and for $a_{20} > a_{19} = 20$ so for $20= a_{a_20}>a_{20} > a_{19} =20$; the only option for such small growth is $d=1$ and $a_k = k + 1$ and so $a_20 = 21$ and $a_{21} = 22$ and $a_{2019} = 2020$. 
