compound interest on a certain sum at a certain rate of interest for the 2nd year is rs. 2200 and for the 3rd year is rs. 2400 find the principal and the rate of interest

Let the Principal be P and rate of interest be r.
Given,
2200 = $$P(1 + r/100)^2 - P(1 + r/100) $$ = P(1 + r/100)(r/100) - (1)
2400 = $$P(1 + r/100)^3 - P(1 + r/100)^2 $$ = P(1 + r/100)^2(r/100) - (2)
(2)$$\div$$1 = 12/11 = 1 + r/100
r = 100/11
Substituting r in (1): P( 1 + 1/11)(1/11) = 2200
P(12/11)(1/11) = 2200
thus, P = Rs 22,183$$\frac{1}{3}$$