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Mathematics · 2023 · 5 marks
CBSE 2023 · Region 1 · Set 1 · Q35
The ratio of the $\displaystyle 11^{\text {th }}$ term to $\displaystyle 17^{\text {th }}$ term of an A.P. is $\displaystyle 3$ : 4. Find the ratio of $\displaystyle 5^{\text {th }}$ term to $\displaystyle 21^{\text {st }}$ term of the same A.P. Also, find the ratio of the sum of first $\displaystyle 5$ terms to that of first $\displaystyle 21$ terms.$\displaystyle 250$ logs are stacked in the following manner : $\displaystyle 22$ logs in the bottom row, $\displaystyle 21$ in the next row, $\displaystyle 20$ in the row next to it and so on (as shown by an example). In how many rows, are the $\displaystyle 250$ logs placed and how many logs are there in the top row ?
(Example)
The ratio of the $\displaystyle 11^{\text {th }}$ term to $\displaystyle 17^{\text {th }}$ term of an A.P. is $\displaystyle 3$ : 4. Find the ratio of $\displaystyle 5^{\text {th }}$ term to $\displaystyle 21^{\text {st }}$ term of the same A.P. Also, find the ratio of the sum of first $\displaystyle 5$ terms to that of first $\displaystyle 21$ terms.
$\displaystyle 250$ logs are stacked in the following manner : $\displaystyle 22$ logs in the bottom row, $\displaystyle 21$ in the next row, $\displaystyle 20$ in the row next to it and so on (as shown by an example). In how many rows, are the $\displaystyle 250$ logs placed and how many logs are there in the top row ?
(Example)
Marking-scheme solution
Given \(\displaystyle \frac{\mathrm{a}+10 \mathrm{~d}}{\mathrm{a}+16 \mathrm{~d}}=\frac{3}{4}\)
\[\begin{aligned}
& \Rightarrow 4 a+40 d=3 a+48 d \\
& \Rightarrow a=8 d
\end{aligned}
\] therefore \(\displaystyle \frac{\mathrm{a}_{5}}{\mathrm{a}_{21}}=\frac{\mathrm{a}+4 \mathrm{~d}}{\mathrm{a}+20 \mathrm{~d}}=\frac{3}{7} \quad\) using( i) \(\displaystyle \mathrm{a}_{5}: \mathrm{a}_{21}=3: 7\)
\[\frac{s_{5}}{s_{21}}=\frac{\dfrac{5}{2}(2 a+4 d)}{\dfrac{21}{2}(2 a+20 d)}=\frac{5 \times 20 d}{21 \times 36 d}=\frac{25}{189}
\] Therefore, \(\displaystyle \mathrm{S}_{5}: \mathrm{S}_{21}=25: 189\)
Let the number of rows be n.
A.P. formed is $\displaystyle 22$, $\displaystyle 21$, $\displaystyle 20$, 19.........
Here \(\displaystyle \mathrm{a}=22, \mathrm{~d}=-1 \mathrm{Sn}=250\)
\[\begin{aligned}
& \therefore 250=\frac{n}{2}[44+(n-1)(-1)] \\
& \Rightarrow n^{2}-45 n+500=0 \\
& \Rightarrow(n-25)(n-20)=0 \\
& n \neq 25 \therefore n=20
\end{aligned}
\]
logs in top row \(\displaystyle =\mathrm{a}_{20}=22+19(-1)=3\)
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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.