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Arithmetic Progressions
CH05 Question Bank
CBSE_CH05_QUESTION_BANK
Grade 10
Question:
[Case Study]
A theatre has 27 rows of seats. Row 1 (nearest the stage) has 20 seats, and each successive row has 2 more seats than the previous row, as shown in the chart for the first six rows.
(a) Find the number of seats in the 15th row. [1 Mark]
(b) Find the number of seats in the last (27th) row. [1 Mark]
(c) Find the total seating capacity of the theatre. [1 Mark]
(d) A latecomer is told that row number $k$ has exactly 58 seats. Find $k$, or explain why no such row exists. [1 Mark]
A theatre has 27 rows of seats. Row 1 (nearest the stage) has 20 seats, and each successive row has 2 more seats than the previous row, as shown in the chart for the first six rows.
(a) Find the number of seats in the 15th row. [1 Mark]
(b) Find the number of seats in the last (27th) row. [1 Mark]
(c) Find the total seating capacity of the theatre. [1 Mark]
(d) A latecomer is told that row number $k$ has exactly 58 seats. Find $k$, or explain why no such row exists. [1 Mark]
Step-by-Step Solution
Key Concept: Case study on arithmetic progressions.
(a) Find the number of seats in the 15th row. [1 Mark]
$a=20,d=2$. $a_{15}=20+14(2)=20+28=48$ seats. [1.0 Mark]
(b) Find the number of seats in the last (27th) row. [1 Mark]
$a_{27}=20+26(2)=20+52=72$ seats. [1.0 Mark]
(c) Find the total seating capacity of the theatre. [1 Mark]
$S_{27}=\dfrac{27}{2}(a+a_{27})=\dfrac{27}{2}(20+72)=\dfrac{27}{2}\times92=27\times46=1242$ seats. [1.0 Mark]
(d) A latecomer is told that row number $k$ has exactly 58 seats. Find $k$, or explain why no such row exists. [1 Mark]
$20+(k-1)2=58\Rightarrow(k-1)2=38\Rightarrow k-1=19\Rightarrow k=20$. Since $1\leq20\leq27$, row $20$ indeed has $58$ seats. [1.0 Mark]
$a=20,d=2$. $a_{15}=20+14(2)=20+28=48$ seats. [1.0 Mark]
(b) Find the number of seats in the last (27th) row. [1 Mark]
$a_{27}=20+26(2)=20+52=72$ seats. [1.0 Mark]
(c) Find the total seating capacity of the theatre. [1 Mark]
$S_{27}=\dfrac{27}{2}(a+a_{27})=\dfrac{27}{2}(20+72)=\dfrac{27}{2}\times92=27\times46=1242$ seats. [1.0 Mark]
(d) A latecomer is told that row number $k$ has exactly 58 seats. Find $k$, or explain why no such row exists. [1 Mark]
$20+(k-1)2=58\Rightarrow(k-1)2=38\Rightarrow k-1=19\Rightarrow k=20$. Since $1\leq20\leq27$, row $20$ indeed has $58$ seats. [1.0 Mark]
Correct Answer: $a=20,d=2$. $a_{15}=20+14(2)=20+28=48$ seats. [1.0 Mark] | $a_{27}=20+26(2)=20+52=72$ seats. [1.0 Mark] | $S_{27}=\dfrac{27}{2}(a+a_{27})=\dfrac{27}{2}(20+72)=\dfrac{27}{2}\times92=27\times46=1242$ seats. [1.0 Mark] | $20+(k-1)2=58\Rightarrow(k-1)2=38\Rightarrow k-1=19\Rightarrow k=20$. Since $1\leq20\leq27$, row $20$ indeed has $58$ seats. [1.0 Mark]
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