The sum of all 3 -digit numbers less than or equal to 500, that are...

$209, 220, 231, . . . . . . . . , 495$ series in A.P.

Clearly,~$a = 209, d = 220 - 209 = 11$~and& $l = 495$

Where~$a$~is the first term,& $d$~is the common difference and& $l$~is the last term.

$⇒ l = a + \left(n - 1\right)d = 495$

where~$n$~is the number of terms

$⇒ l = 209 + \left(n - 1\right)11 = 495$

$⇒ \left(n - 1\right)11 = 286$

$⇒ \left(n - 1\right) = 26$

$⇒ n = 27$

Sum$= \dfrac{27}{2}\left(2 \times 209 + \left(26\right) \times 11\right)= 27\left(209 + 143\right) = 27 \times 352 = 9504$

Numbers containing~$1$~at unit place $= 231, 341, 451$

Numbers containing $1$~at ${10}^{th}$~place $= 319, 418$

Hence, required sum of all~$3$-digit~numbers less than or equal to& $500= 9501 - \left(231 + 341 + 451 + 319 + 418\right) = 7744$