Are the total number of electrons present in the central atom of triiodide equal to 10? My book states that 


  the total number of electrons present in the central atom of $\ce{I_3^-}$ is equal to 10. 


What if I put a negative charge on the central iodine atom such that it gains a noble configuration and then I say that it is coordinatively bonded with other two iodine atoms? Then the number of electrons which the central atom will have would be 8. Wouldn't it? 

Am I wrong somewhere in my analogy?
Looking at the wikipedia article here, your explanation seems to be exactly right:


  The ion is linear and symmetrical. According to VSEPR theory, the central iodine atom has three equatorial lone pairs, and the terminal iodines are bonded axially in a linear fashion, due to the three lone pairs bonding to the central iodine-atom. In the molecular orbital model, a common explanation for the hypervalent bonding on the central iodine involves a three-center four-electron bond.


Thus, the central iodine atom has $3\cdot 2$ valence electrons from the lone pairs and 3 three-center four-electron bonds, so that I speculate you end up with $3\cdot 2$ VE + $4/3 \cdot 3$ VE = 6 VE + 4 VE = 10 VE. Which also fits to the reference given by you.