JEE Main 2021 (Online) 3rd August Evening ShiftQuestion 1
If the sum of all $3$-digit numbers which are multiples of $9$ is $41k$, then '$k$' is equal to $\_\_\_\_\_\_$ .
Sequence and Series · Arithmetic Progression
Answer$1350$
Solution
Formulas used. (1) $n=\frac{l-a}{d}+1$ and $S_n=\frac n2\,(a+l)$Step 1 - The A.P. The smallest $3$-digit multiple of $9$ is $108$ and the largest is $999$: $$n=\frac{999-108}{9}+1=100 .$$ Step 2 - Sum. $$\frac{100}{2}(108+999)=50\times1107=55350 .$$ Step 3. $$41k=55350\ \Longrightarrow\ k=1350 .$$ Answer: 1350