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Answer :
(i)Taxi fare for 1st km = Rs 15,
Taxi fare after 2 km = 15 + 8 = Rs 23
Taxi fare after 3 km = 23 + 8 = Rs 31
Taxi fare after 4 km = 31 + 8 = Rs 39
Therefore, the sequence is 15, 23, 31, 39…
It is an arithmetic progression because difference between any two consecutive terms is equal which is 8.
(23 – 15 = 8, 31 – 23 = 8, 39 – 31 = 8, …)
(ii)Let amount of air initially present in a cylinder = V
Amount of air left after pumping out air by vacuum pump
= \( V - {{V} \over {4}} \)
\( = {{4V - V} \over {4}} \)
\( = {{3V} \over {4}}\)
Amount of air left when vacuum pump again pumps out air
=\({{3V} \over {4}} - ( {{1} \over {4}} × {{3V} \over {4}} ) \)
\( = {{3V} \over {4}} - {{3V} \over {16}} \)
\( = {{12V - 3V} \over {16}} \)
\( = {{9V} \over {16}}\)
So, the sequence we get is like \(V, {{3V} \over {4}}, {{9V} \over {16}}.....\)
Checking for difference between consecutive terms …
\({{3V} \over {4}} - V = {{-V} \over {4}}\),
\({{9V} \over {16}} - {{3V} \over {4}} = {{9V - 12V} \over {16}} = {{-3V} \over {16}}\)
Difference between consecutive terms is not equal.
Therefore, it is not an arithmetic progression.
(iii) Cost of digging 1 meter of well = Rs 150
Cost of digging 2 meters of well = 150 + 50 = Rs 200
Cost of digging 3 meters of well = 200 + 50 = Rs 250
Therefore, we get a sequence of the form 150, 200, 250…
It is an arithmetic progression because difference between any two consecutive terms is equal. (200 – 150 = 250 – 200 = 50…)
Here, difference between any two consecutive terms (common difference) is equal to 50.
(iv)Amount in bank after Ist year = \(1000 ( 1 + {{8} \over {100}})\)… (1)
Amount in bank after two years = \(1000 ( 1 + {{8} \over {100}})^2\)… (2)
Amount in bank after three years = \(1000 ( 1 + {{8} \over {100}})^3\)… (3)
Amount in bank after four years = \(1000 ( 1 + {{8} \over {100}})^4\)… (4)
It is not an arithmetic progression because \((2) - (1) \ne (3) - (2)\)
(Difference between consecutive terms is not equal)
Therefore, it is not an Arithmetic Progression.