What is significance of memory at 0000:7c00 to booting sequence?

The ":" is a holdover from segmented memory days, when PCs ran in real mode and could only do 64K at a time. The number to the left of the ":" is your segment, the number to the right is your address.

The windows debug command accepts this notation if you want to poke around in memory yourself:

C:\Users\Seth> debug
-d0000:7c00
0000:7C00  00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00   ................
0000:7C10  00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00   ................
0000:7C20  00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00   ................
0000:7C30  00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00   ................
0000:7C40  00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00   ................
0000:7C50  00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00   ................
0000:7C60  00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00   ................
0000:7C70  00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00   ................

With regard to this particular address, it's just an address that was picked to load the MBR, See: https://web.archive.org/web/20140701052540/http://www.ata-atapi.com/hiwmbr.html

"If an MBR is found it is read into memory at location 0000:7c00 and INT 19 jumps to memory location 0000:7c00"

In the original IBM PC it was thought inconceivable to have more than 32K RAM. In segmented addressing terms this is 0000:8000 where 8000 hex is 32768 decimal. The fashion of the time was for the BIOS POST conclude by loading the Boot Sector of the floppy in A: or the Master Boot Record of the hard drive in C: at the location 512 bytes below the top of memory which means subtract 0200 hex from 8000 hex to get 7C00. So the boot sequence loaded the first valid 512 byte first sector into, and then set the Instruction Pointer to 0000:7C00 to execute it. I used to write the code for these first sectors to load the operating system.