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Arithmetic Progressions
CH05 Question Bank
CBSE_CH05_QUESTION_BANK
Grade 10
Question:
In an auditorium, seats are arranged in rows as shown. Row 1 has 20 seats, row 2 has 22 seats, row 3 has 24 seats, and so on, with each subsequent row having 2 more seats than the previous one. If there are 25 rows in total, find (i) the number of seats in the last row, and (ii) the total seating capacity of the auditorium.
Step-by-Step Solution
Key Concept: Recognise the row-wise seat counts as an AP; apply the nth-term formula for part (i) and the sum formula for part (ii).
The seat counts form an AP with $a=20,d=2,n=25$. [1.0 Mark]
(i) $a_{25}=20+24(2)=20+48=68$ seats in the last row. [1.5 Marks]
(ii) $S_{25}=\dfrac{25}{2}(a+a_{25})=\dfrac{25}{2}(20+68)=\dfrac{25}{2}\times88$. [1.5 Marks]
$=25\times44=1100$ seats in total. [1.0 Mark]
(i) $a_{25}=20+24(2)=20+48=68$ seats in the last row. [1.5 Marks]
(ii) $S_{25}=\dfrac{25}{2}(a+a_{25})=\dfrac{25}{2}(20+68)=\dfrac{25}{2}\times88$. [1.5 Marks]
$=25\times44=1100$ seats in total. [1.0 Mark]
Correct Answer:
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