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+(n−1)d=495
where n is the number of terms
⇒l=209+(n−1)11=495
⇒(n−1)11=286
⇒(n−1)=26
⇒n=27
Sum=227(2×209+(26)×11)=27(209+143)=27×352=9504
Numbers containing 1 at unit place =231,341,451
Numbers containing 1 at 10th place =319,418
Hence, required sum of all 3-digit numbers less than or equal to 500=9501−(231+341+451+319+418)=7744