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Mathematics · 2024 · 5 marks
CBSE 2024 · Region 1 · Set 1 · Q33
(A) The sum of first and eighth terms of an A.P. is 32 and their product is 60. Find the first term and common difference of the A.P. Hence, also…
(A)
The sum of first and eighth terms of an A.P. is $\displaystyle 32$ and their product is $\displaystyle 60$ . Find the first term and common difference of the A.P. Hence, also find the sum of its first $\displaystyle 20$ terms.
OR
(B)
In an A.P. of $\displaystyle 40$ terms, the sum of first $\displaystyle 9$ terms is $\displaystyle 153$ and the sum of last $\displaystyle 6$ terms is $\displaystyle 687$ . Determine the first term and common difference of A.P. Also, find the sum of all the terms of the A.P.
Marking-scheme solution
\[\begin{aligned}
& a+a_{8}=32 \Longrightarrow 2 a+7 d=32 \\
& a \times a_{8}=60 \Longrightarrow a(a+7 d)=60
\end{aligned}
\]
Solving (i) & (ii), we get
\[\mathrm{a}=2 \text { or } \mathrm{a}=30
\]
and \(\displaystyle \mathrm{d}=4\) or \(\displaystyle \mathrm{d}=-4\)
First term and common difference of A.P. are $\displaystyle 2$ and $\displaystyle 4$ or $\displaystyle 30$ and - $\displaystyle 4$ respectively.
Now, for \(\displaystyle \mathrm{a}=2\) & \(\displaystyle \mathrm{d}=4\)
\[S_{20}=10(4+76)=800
\]
and for \(\displaystyle \mathrm{a}=30 \& \mathrm{~d}=-4\)
\[S_{20}=10(60-76)=-160
\]
Here \(\displaystyle \mathrm{n}=40\),
\[\mathrm{S}_{9}=\frac{9}{2}[2 a+8 d]=153 \Rightarrow \mathrm{a}+4 \mathrm{~d}=17
\]
and \(\displaystyle \mathrm{S}_{40}-\mathrm{S}_{34}=687\) or \(\displaystyle \mathrm{a}_{35}+\mathrm{a}_{36}+\mathrm{a}_{37}+\mathrm{a}_{38}+\mathrm{a}_{39}+\mathrm{a}_{40}=687\)
\[\Rightarrow 6 \mathrm{a}+219 \mathrm{~d}=687 \text { or } 2 \mathrm{a}+73 \mathrm{~d}=229
\]
solving (i) and (ii) to get \(\displaystyle \mathrm{a}=5, \mathrm{~d}=3\)
\[\text { Also, } S_{40}=\frac{40}{2}(10+39 \times 3)=2540
\]
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CBSE Class 10 Mathematics past-paper question from the 2024 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.