Voltage shifting and dividing
Since you need to attenuate the signal, the conversion of the input &plusmn;10 V signal to 0-2.5 V output can be accomplished with:



R1 and R2 form a voltage divider.  This divider sets the gain from input to output.  V1 adds a arbitrary offset.

Start by solving the voltage divider.  The input range is 20 Vpp, and the output 2.5 Vpp.  That means the divider must have a gain of 2.5/20 = 1/8.  That means R1 = 7 R2, or conversely, R2 = R1 / 7.  Note that this is only based on the input to output amplitudes, and has nothing to do with the offset.

With the R1 and R2 ratio set, the voltage source value can be found by solving at any one operating point.  You can choose any convenient point.  I'll use -10 V in.  With -10 V in, we know the output will be 0 V.  The voltage across each resistor is proportional to its resistance.  From inspection, we can see that there is 10 V across R1, so there must be 10/7 V across R2.  V1 must be positive to result in 0 V out when the input is negative.  V1 must therefore be 10/7 V = 1.43 V.

So far we've only fixed the ratio between R1 and R2.  We have another degree of freedom while still getting the specified result.  That freedom sets the impedance of the divider.  You haven't said what impedance the &plusmn;10 V source can drive, and what impedance the A/D input requires.  Since there is no spec here, I'll arbitrarily pick 20 k&Omega; for R1.  That means R2 must be 2.86 k&Omega;.  The input impedance the &plusmn;10 V source sees is therefore 22.9 k&Omega;, and the output impedance the A/D is driven with is 2.5 k&Omega;.  You can scale both R1 and R2 by the same amount to adjust these values.

While you could create a 1.43 V source to drive the bottom end of R2, a simpler method results from thinking of R2 and V1 together as a Thevenin source.  Let's say you have a 3.3 V supply available.  With two resistors between this supply and ground, you can create a Thevenin source of 1.43 V and 2.86 k&Omega;.  The final circuit then becomes:



R1 is the same as before.  R3, R4, and the 3.3 V supply together replace R2 and V1 in the previous circuit.  R3//R4 must be R2, and just R3 and R4 by themselves dividing the 3.3 V supply must result in 1.43 V.

So after a bunch of analysis, the answer is just three resistors.  Yes, it really is that simple.