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A tent house owner provides furniture on rent. He stacks chairs in his shop to save space.
In the diagram, the height of seat of chair from ground is represented by $h_1, h_2, h_3, ....$. The height of first seat is $44$ cm from ground level and gap between every two seats is $10$ cm.
(i) Write the values of $h_1, h_2, h_3, h_4$ and $h_5$ in this order only.
(ii) Show that the above values form an A.P. Write its first term and common difference.
(iii) (a) If chairs can be stacked up to the maximum height of $160$ cm, then find the maximum number of chairs in a stack.
OR
(iii) (b) Is it possible to stack $15$ chairs if maximum height of the stack can not be more than $180$ cm? Justify your answer.
In the diagram, the height of seat of chair from ground is represented by $h_1, h_2, h_3, ....$. The height of first seat is $44$ cm from ground level and gap between every two seats is $10$ cm.
(i) Write the values of $h_1, h_2, h_3, h_4$ and $h_5$ in this order only.
(ii) Show that the above values form an A.P. Write its first term and common difference.
(iii) (a) If chairs can be stacked up to the maximum height of $160$ cm, then find the maximum number of chairs in a stack.
OR
(iii) (b) Is it possible to stack $15$ chairs if maximum height of the stack can not be more than $180$ cm? Justify your answer.
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Solution: (i) $h_1 = 44, h_2 = 54, h_3 = 64, h_4 = 74, h_5 = 84$
(ii) Since gap between heights of seats of every two adjacent chairs is same $\therefore h_1, h_2, h_3, .....$ form an A.P. Here, $a = 44$ and $d = 10$
(iii) (a) $160 = 44 + (n - 1) \times 10 \Rightarrow n = 12.6 \therefore$ maximum $12$ chairs can be stacked up.
OR
(iii) (b) $h_{15} = 44 + 14 \times 10 = 184$ cm. $184$ cm $> 180$ cm. $\therefore 15$ chairs cannot be stacked up.
(ii) Since gap between heights of seats of every two adjacent chairs is same $\therefore h_1, h_2, h_3, .....$ form an A.P. Here, $a = 44$ and $d = 10$
(iii) (a) $160 = 44 + (n - 1) \times 10 \Rightarrow n = 12.6 \therefore$ maximum $12$ chairs can be stacked up.
OR
(iii) (b) $h_{15} = 44 + 14 \times 10 = 184$ cm. $184$ cm $> 180$ cm. $\therefore 15$ chairs cannot be stacked up.