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Mathematics · 2024 · 3 marks
CBSE 2024 · Region 3 · Set 1 · Q26
If the sum of first $\displaystyle 7$ terms of an A.P. is $\displaystyle 49$ and that of first $\displaystyle 17$ terms is $\displaystyle 289$, find the sum of its first $\displaystyle 20$ terms.The ratio of the $\displaystyle 10^{\text {th }}$ term to its $\displaystyle 30^{\text {th }}$ term of an A.P. is $\displaystyle 1$ : $\displaystyle 3$ and the sum of its first six terms is 42. Find the first term and the common difference of A.P.
If the sum of first $\displaystyle 7$ terms of an A.P. is $\displaystyle 49$ and that of first $\displaystyle 17$ terms is $\displaystyle 289$, find the sum of its first $\displaystyle 20$ terms.
The ratio of the $\displaystyle 10^{\text {th }}$ term to its $\displaystyle 30^{\text {th }}$ term of an A.P. is $\displaystyle 1$ : $\displaystyle 3$ and the sum of its first six terms is 42. Find the first term and the common difference of A.P.
Marking-scheme solution
Let a be the first term and d be the common difference.
\[\begin{aligned}
& \frac{7}{2}(2 \mathrm{a}+6 \mathrm{~d})=49 \\
& \mathrm{a}+3 \mathrm{~d}=7 \ldots \ldots \ldots \ldots \ldots \\
& S_{17}=289 \\
& \frac{17}{2}(2 \mathrm{a}+16 \mathrm{~d})=289 \\
& \mathrm{a}+8 \mathrm{~d}=17 \ldots \ldots \ldots \ldots .
\end{aligned}
\] (ii) solving (i) and (ii)
\[\begin{aligned}
\mathrm{d} & =2 \& \mathrm{a}=1 \\
S_{20} & =\frac{20}{2}[2(1)+19(2)] \\
& =400
\end{aligned}
\]
Let a be the first term and d be the common difference.
\[\begin{aligned}
& \frac{a+9 d}{a+29 d}=\frac{1}{3} \\
& \Rightarrow \mathrm{a}=\mathrm{d}
\end{aligned}
\]
\[\begin{aligned}
& \frac{6}{2}(2 a+5 d)=42 \\
& \Rightarrow 2 a+5 d=14
\end{aligned}
\] Solving (i) and (ii)
\(\displaystyle \mathrm{a}=2\) and \(\displaystyle \mathrm{d}=2\)
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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.