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Mathematics · 2024 · 3 marks
CBSE 2024 · Region 2 · Set 1 · Q26
If the sum of first m terms of an A.P. is same as sum of its first n terms $\displaystyle (\mathrm{m} \neq \mathrm{n})$, then show that the sum of its first $\displaystyle (\mathrm{m}+\mathrm{n})$ terms is zero.In an A.P., the sum of three consecutive terms is $\displaystyle 24$ and the sum of their squares is $\displaystyle 194$ . Find the numbers.
If the sum of first m terms of an A.P. is same as sum of its first n terms $\displaystyle (\mathrm{m} \neq \mathrm{n})$, then show that the sum of its first $\displaystyle (\mathrm{m}+\mathrm{n})$ terms is zero.
In an A.P., the sum of three consecutive terms is $\displaystyle 24$ and the sum of their squares is $\displaystyle 194$ . Find the numbers.
Marking-scheme solution
\[\begin{aligned}
& S_{m}=S_{n} \\
& \Rightarrow \frac{m}{2}\left[2 a+(m-1) d=\frac{n}{2}[2 a+(n-1) d]\right. \\
& \Rightarrow 2 a(m-n)=d\left(n^{2}-m^{2}\right)-d(n-m) \\
& \Rightarrow 2 a=-d(m+n-1) \\
& \text { or } 2 a+(m+n-1) d=0 \\
& \text { i.e., } S_{m+n}=\frac{m+n}{2}[2 a+(m+n-1) d]=0
\end{aligned}
\]
Let the numbers be a - d, a, a + d
\[\begin{aligned}
& \therefore a-d+a+a+d=24 \\
& \Rightarrow a=8
\end{aligned}
\]
Also, \(\displaystyle (\mathrm{a}-\mathrm{d})^{2}+\mathrm{a}^{2}+(\mathrm{a}+\mathrm{d})^{2}=194\)
\[\begin{aligned}
& \Rightarrow(8-d)^{2}+8^{2}+(8+d)^{2}=194 \\
& \Rightarrow d^{2}=1 \Rightarrow d= \pm 1 \\
& \therefore \text { Numbers are } 7,8,9 \text { or } 9,8,7
\end{aligned}
\]
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CBSE Class 10 Mathematics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.