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Mathematics · 2026 · 2 marks
CBSE 2026 · Region 3 · Set 3 · Q23
In an A.P., the first term is $\displaystyle 4$ and the last term is 31. If sum of all the terms is $\displaystyle 175$, find the number of terms and the common difference.How many terms of the A.P. $\displaystyle 21$, $\displaystyle 18$, $\displaystyle 15$, .... must be added to get the sum zero ?
In an A.P., the first term is $\displaystyle 4$ and the last term is 31. If sum of all the terms is $\displaystyle 175$, find the number of terms and the common difference.
How many terms of the A.P. $\displaystyle 21$, $\displaystyle 18$, $\displaystyle 15$, .... must be added to get the sum zero ?
Marking-scheme solution
Here \(\displaystyle \mathrm{a}=4, l=31\)
\[\begin{aligned}
& S_{n}=175 \\
\Rightarrow & \frac{n}{2}(4+31)=175 \\
\Rightarrow & n=10
\end{aligned}
\]
So, \(\displaystyle \mathrm{a}_{10}=31\)
\[\begin{aligned}
& \Rightarrow 4+9 d=31 \\
& \Rightarrow d=3
\end{aligned}
\]
\[\begin{aligned}
d & =18-21=-3 \\
& S_{n}=0 \\
\Rightarrow & \frac{n}{2}[2 \times 21+(n-1) \times(-3)]=0 \\
\Rightarrow & n^{2}-15 n=0 \Rightarrow n(n-15)=0 \\
\Rightarrow & n=0,15
\end{aligned}
\]
Since 'n' is a natural number.
\[\therefore \mathrm{n}=15
\]
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CBSE Class 10 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.