CBSE 2025 · Region 3 · Set 1 · Q33 · 5 marks
The sum of the third term and the seventh term of an AP is $\displaystyle 6$ and their product is 8. Find the sum of the first sixteen terms of the AP.The minimum age of children eligible to participate in a painting competition is $\displaystyle 8$ years. It is observed that the age of the youngest boy was $\displaystyle 8$ years and the ages of the participants, when seated in order of age, have a common difference of $\displaystyle 4$ months. If the sum of the ages of all the participants is $\displaystyle 168$ years, find the age of the eldest participant in the painting competition.
The sum of the third term and the seventh term of an AP is $\displaystyle 6$ and their product is 8. Find the sum of the first sixteen terms of the AP.
The minimum age of children eligible to participate in a painting competition is $\displaystyle 8$ years. It is observed that the age of the youngest boy was $\displaystyle 8$ years and the ages of the participants, when seated in order of age, have a common difference of $\displaystyle 4$ months. If the sum of the ages of all the participants is $\displaystyle 168$ years, find the age of the eldest participant in the painting competition.
Marking-scheme solution
(a)
Let first term = a and common difference = d
ATQ, \(\displaystyle (\mathrm{a}+2 \mathrm{~d})+(\mathrm{a}+6 \mathrm{~d})=6\)
\(\displaystyle \mathrm{a}+4 \mathrm{~d}=3\)
\(\displaystyle \mathrm{a}=3-4 \mathrm{~d}\)
Also, \(\displaystyle (\mathrm{a}+2 \mathrm{~d})(\mathrm{a}+6 \mathrm{~d})=8\)
\(\displaystyle (3-4 d+2 d)(3-4 d+6 d)=8\)
\(\displaystyle 9-4 \mathrm{~d}^{2}=8\)
\(\displaystyle \mathrm{d}= \pm \frac{1}{2}\)
When \(\displaystyle \mathrm{d}=\frac{1}{2} \Rightarrow \mathrm{a}=1\)
\(\displaystyle \mathrm{S}_{16}=\frac{16}{2}\left[2 \times 1+15 \times \frac{1}{2}\right]\)
\(\displaystyle =76\)When \(\displaystyle \mathrm{d}=-\frac{1}{2} \Rightarrow \mathrm{a}=5\)
\(\displaystyle \mathrm{S}_{16}=\frac{16}{2}\left[2 \times 5+15 \times\left(-\frac{1}{2}\right)\right]\)
\(\displaystyle =20\)
OR
(b)The ages of the participants form the following AP\(\displaystyle 8,8 \frac{1}{3}, 8 \frac{2}{3}, 9, \cdots\)
where first term \(\displaystyle =8\) and common difference \(\displaystyle =\frac{1}{3}\)
Let the number of participants be n
\(\displaystyle \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2 \times 8+(\mathrm{n}-1) \frac{1}{3}\right]=168\)
\(\displaystyle \mathrm{n}^{2}+47 \mathrm{n}-1008=0\)
\(\displaystyle \Rightarrow \mathrm{n}=16\)
\(\displaystyle \therefore\) the age of the eldest participant \(\displaystyle =8+15 \times \frac{1}{3}=13\) years
Arithmetic ProgressionsSum of First n Terms of an APApplylong_answermedium
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CBSE Class 10 Mathematics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.