A twenty storied building was observed from outside, each floor has a kite shaped figure which gets on magnifying as observed from the $1^{s t}$ to $20^{t h}$ floor. The area of first five kites on each floor was calculated and they were found to develop a pattern. These five areas are ${10 m}^{2}$ , ${15 m}^{2}$ , ${20 m}^{2}$ , ${25 m}^{2}$ , ${30 m}^{2}$ and so on. The area of the kite on the top floor is ____.

${100 m}^{2}$

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${105 m}^{2}$

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${110 m}^{2}$

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${200 m}^{2}$

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Solution

The correct option is B
${105 m}^{2}$

The sequence of the areas on each floor forms an Arithmetic Progression, with first term as 10 and common difference as 5.

General term $T_{n} = a + (n - 1) d$
$= 10 + (n - 1) 5$

We require, $T_{20} = 10 + (20 - 1) 5 = 10 + 19 \times 5 = 105$

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A twenty storied building was observed from outside, each floor has a kite shaped figure which gets on magnifying as observed from the $1^{s t}$ to $20^{t h}$ floor. The area of first five kites on each floor was calculated and they were found to develop a pattern. These five areas are ${10 m}^{2}$ , ${15 m}^{2}$ , ${20 m}^{2}$ , ${25 m}^{2}$ , ${30 m}^{2}$ and so on. The area of the kite on the top floor is ____.

Q. A barometer is carried from the

1^{s t}

floor to the

20^{t h}

floor of a building. Why does the reading on the barometer fall?

On which floor of the building does G stay? (the building has five floors 1, 2, 3, 4, 5. )
I. Only the even-numbered floors are occupied and G does not stay on the second floor.
II. G does not stay on an odd-numbered floor.

Q. A barometer is carried from a first floor 20th floor of a building why does the reading on the Barometer fall

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The correct option is B 105m2 The sequence of the areas on each floor forms an Arithmetic Progression, with first term as 10 and common difference as 5. General term Tn=a+(n−1)d =10+(n−1)5 We require,T20=10+(20−1)5=10+19×5=105

The correct option is B
${105 m}^{2}$

The sequence of the areas on each floor forms an Arithmetic Progression, with first term as 10 and common difference as 5.

General term $T_{n} = a + (n - 1) d$
$= 10 + (n - 1) 5$

We require, $T_{20} = 10 + (20 - 1) 5 = 10 + 19 \times 5 = 105$