Using source command in script but defining the input file in the command line of a terminal

From help source:

source: source filename [arguments]
    Execute commands from a file in the current shell.

    Read and execute commands from FILENAME in the current shell.  The
    entries in $PATH are used to find the directory containing FILENAME.
    If any ARGUMENTS are supplied, they become the positional parameters
    when FILENAME is executed.

    Exit Status:
    Returns the status of the last command executed in FILENAME; fails if
    FILENAME cannot be read.

So, in the script, you only need to add this line:

source input.in

or this one (the POSIX version):

. input.in

To pass the input file at runtime, you use positional parameters:

source "$1"

. "$1"

Also note that d=a*b+c won't work unless d has the "integer" attribute:

declare -i d
d=a*b+c

or you use arithmetic expansion to perform the operation:

d=$((a*b+c))

Example:

#!/bin/bash
source "$1"
d=$((a*b+c))
echo "$d"

$ ./script.sh input.in
170

This is a very basic operation. Just source or . a file name, and the result is as if those lines were included in your primary script.

The syntax is simply this:

source file_name

or

. file_name

There are some nuances, but that's more advanced. See man bash | grep source for more details.