Mathematics - Permutation & Combination Question with Solution | TestHub

$209,220,231,........,495$ 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.

$\Rightarrow l=a+\left(n-1\right)d=495$

where $n$ is the number of terms

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

$\Rightarrow \left(n-1\right)11=286$

$\Rightarrow \left(n-1\right)=26$

$\Rightarrow n=27$

Sum$=\frac{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 ${10}^{\text{th}}$ 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$