CBSE 2025 · Region 2 · Set 3 · Q27 · 3 marks
Find the sum of all $\displaystyle 3$-digit natural numbers which are divisible by 11.
Marking-scheme solution
$\displaystyle 3$ - digit natural numbers divisible by $\displaystyle 11$ are
\[110,121, \ldots, 990
\]
Here first term = $\displaystyle 110$ and common difference = $\displaystyle 11$
\[\begin{aligned}
& a_{n}=990 \\
\Rightarrow & 110+(n-1) \times 11=990 \\
\Rightarrow & n=81 \\
S_{81}= & \frac{81}{2} \times[110+990] \\
= & 44550
\end{aligned}
\]
Arithmetic ProgressionsSum of First n Terms of an APApplyshort_answermedium
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CBSE Class 10 Mathematics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.