Arithmetic Sequence coefficients

$f(f(x))=b(bx+a)+a=b^2 x+ ab+a$ $\therefore c=ab+a, d=b^2$ Using condition for them being in arithmetic progression:- $b-a=ab+a-b=b^2-ab-a$ $\therefore b-a=\frac{ab}2, 2a(b+1)=(b+1)b$ $b+1=0$ or $2a=b$

If $b=-1$ the series turns out to be $\{-2,-1,0,1\}$

If $2a=b$ it comes as $\{1,2,3,4\}$

Thus $S=4026+1-2013-2=2012$

$ff(x) = b(bx +a) + a = dx +c$ $b^2x + ab + a = dx + c$ $b^2 = d ... (1), ab+a = c ... (2)$ Since $a, b, c, d$ are consecutive terms in an arithmetic sequence, lets denote $a = A, b = A + D, c = A+2D, d = A+3D$ . From (2), $A(A+D) + A = A+ 2D$ $A^2 + AD -2D = 0$ $A = -\frac{D}{2} \pm \frac {\sqrt {D^2 + 8D}}{2}$ Since A must be an integer, $D^2 + 8D$ must be a perfect square. Let $D^2 + 8D = k^2$ where k is an integer, $D = -4 \pm \sqrt { 4^2 + k^2 }$ Since D must also be an integer, $4^2 + k^2$ must be a perfect square. From Pythagoras Theorem, the only value of k which fulfills is when $k =3$ . (Set of $(3,4,5)$ ) $\Rightarrow D = -4 \pm 5$ $D = 1, -9$ For $D = 1$ , $A = -\frac {1}{2} \pm \frac {\sqrt {9}}{2}$ $A = 1, -2$ For $D = -9$ , $A = \frac {9}{2} \pm \frac {\sqrt {9}}{2}$ $A = 6, 3$ From (1), $(A+D)^2 = A + 3D$ $A^2 + (2D-1)A + D^2 -3D = 0$ $A = \frac {-2D+1}{2} \pm \frac { \sqrt {8D+1}}{2}$ When $D=1$ , $A = -\frac{1}{2} \pm \frac {\sqrt {9}}{2}$ $A = 1, -2$ $\Rightarrow$ True for both equation (1) and (2).

When $D = -9$ , $A = \frac {18 +1}{2} \pm \frac {\sqrt {-71}}{2}$ $\Rightarrow A$ will not be a real number, so $D = -9$ is rejected.

For $D = 1, A = 1$ , $f(x) = 2x + 1$ $f(2013) = 4027$ For $D = 1, A = -2$ , $f(x) = -x - 2$ $f(2013) -2015$ So the sum of all possible values is $4027 + (-2015) = 2012$ where the last three digits are $012$ .

Let a,b,c,d be respectively a,a+d,a+2d,a+3d . Now Replace these values in f(x)=bx+a and f(f(x))=dx+c. Now we will end up getting two equations as $p=\frac{a^{2}}{(2-a)}$ and $a+3p=a+p^{2}$ . Solving these equations, we get a=(-2) or (1) and respectively p=1 . Now we calculate the value of x from * 2013=bx+a * in both cases. Now we put that value of x in * f(f(x))=dx+c * . So we get 2 values as 4027 and (-2015). finally the total value of f(2013) in total is 2012

$f(x)=bx+a \implies f(f(x))=b(bx+a)+a= b^2 x+ab+a$ But according to the question, $b^2 x+ab+a=dx+a$ So, $b^2 =d$ and $(b+1)a=c$ . Let $b=a+m, c=a+2m, d=a+3m$ where $m$ is the common difference.

So, $(b+1)a=c \implies (a+m+1)a=a+2m \implies a^2+am-2m=0$ ....................(1)

$b^2 =d \implies (a+m)^2=a+3m \implies a^2+m^2+2am-a-3m=0$ $\implies (a^2+am-2m)+m^2+am-a-m=0$ [from (1)] $\implies m^2+am-a-m=0 \implies (a+m)(m-1)=0$

So, either $a=-m$ or $m=1$ . But $a=-m$ makes $b=0$ and $d=-2m$ which means $f(x)$ is a constant function as well as a non- constant function which is not true. Even if we try to make the function constant, we end up with $m=0$ which violates the statement $"a,b,c,d$ $are$ $in$ $AP"$ . So, $m=1$ .

Putting $m=1$ in (1) we get
$a^2+a-2=0 \implies (a+2)(a-1)=0 \implies a=-2,1$ . Thus, we get two sets of values for ${a,b,c,d}$ . The sets are ${-2,-1,0,1}$ and ${1,2,3,4}$ .

Hence, $f(x)=-x-2$ or $f(x)=2x+1$ . Putting $x=2013$ in bot functions and adding them, we get the last 3 digits of the result as $012$ .

$f(f(x))=f(bx+a)=b(bx+a)+a=b^2x+ab+a\implies d=b^2 \wedge c=ab+a$

since $a,b,c,d$ is an arithmetic sequence we can represent it as:

$a=a$

$b=a+k$

$c=a+2k$

$d=a+3k$

and this can be written as

$b=a+k$

$ab+a=a+2k\implies ab=2k$

$b^2=a+3k$

by solving this system we get three possibilities:

$a=-2 \wedge b=-1 \wedge k=1\implies f(x)=-x-2 \implies f(2013)=-2015$

$a=0 \wedge b=0 \wedge k=0\implies f(x)=0x+0 \implies f(2013)=0$

$a=1 \wedge b=2 \wedge k=-1\implies f(x)=2x+1 \implies f(2013)=4027$

so the sum of the values of $f(2013)$ is $-2015+0+4027=2\boxed{012}$

Let, a,b,c,d =a,a+e,a+2e,a+3e. f(bx+a) =dx+c implies b bx+ab+a=dx+c. implies b b=d & a(b+1)=c implies (a+e).(a+e)=a+3e &a(a+e+1)=a+2e this gives e=1; a=1,-2; then f(2013)=4027; f(2013)=-2015 then 4027+(-2015)=2012