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Mathematics · 2023 · 5 marks
CBSE 2023 · Region 2 · Set 2 · Q34
Solve the equation for x : \[1+4+7+10+\ldots .+x=287 \]
Marking-scheme solution
\[\begin{aligned}
& 1+4+7+10+\ldots \ldots+x=287 \\
& a=1, d=3 \text { Last term }=x \\
& \Rightarrow 1+(n-1) 3=x \\
& 3 n-2=x \Rightarrow n=\frac{x+2}{3}-(i) \\
& S_{n}=287 \\
& \frac{n}{2}[1+x]=287 \Rightarrow \frac{(x+2)(x+1)}{6}=287-\text { using (i) } \\
& x^{2}+3 x+2=1722 \\
& x^{2}+3 x-1720=0 \\
& \Rightarrow(x+43)(x-40)=0 \\
& \Rightarrow x=-43,40 \\
& x \neq-43 \\
& \therefore x=40
\end{aligned}
\]
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CBSE Class 10 Mathematics past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.