 Hello and welcome to the session. I am Deepika here. Let's discuss the question in the following situation Does the list of numbers involved make an automatic progression and why? Situation is the amount of air present in the cylinder when a vacuum pump Removes 1 by 4 of the air remaining in the cylinder at a time. So let's start the solution initially the volume of air in the cylinder is equal to V is equal to V Let us take this as number one now a vacuum pump removes 1 by 4 of the air in the cylinder So volume of air removed is equal to 1 by 4 of V that is equal to V by 4 Therefore remaining air in the cylinder is initially it was V now air removed is V by 4 So this is equal to 3 by 4 V So let us take this as number 2 that is 3 V by 4. So this is equal to 1 by 4 into 3 V by 4 that is equal to 3 V upon 16 Now therefore the remaining air is is equal to 3 V by 4 3 V by 16 So this is equal to 5 16 V Which is again equal to 3 by 4 square V So let us take this as number 3 by 4 of the remaining air that is 3 by 9 by 16 V is removed So this is equal to 9 by 64 V Remaining air by 64 V is removed is equal to 9 by 16 V minus 9 by 64 V which is equal to 27 by 64 V that is again equal to 3 by 4 Q V So let us take this as number 4 to 3 4 We have the terms as follows First was V then it was 3 by 4 V again 3 by 4 square V and 3 by 4 and so on each successive term is not obtained by adding a fixed number to the preceding term fixed number Because in AP the each term successive term is found by adding a fixed number to the preceding term does not form an AP and our answer is No, the numbers formed according to the situation does not form an AP and volumes are by 4 3 by 4 Square V. Hope the question is clear to you. I and have a nice day