किसी बेलन में उपस्थित हवा की मात्रा, जबकि वायु निकालने वाला पंप प्रत्येक बार बेलन की शेष हवा का 1/4भाग बाहर निकाल देता है।
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Step-by-step explanation:
The amount of air present in a cylinder, while the exhaust pump pumps out 1/4 of the remaining air in the cylinder each time
- Let initially volume of cylinder in the air be a
- So a1 = a
- So a2 = a – ¼ a
- = 3a / 4
- So a3 = ¾ a – ¼ x 3a / 4
- = 3a/4 – 3a / 16
- = 9/16 a
- So a4 = 9/16 a – (9/16 a x ¼)
- 9a / 16 – 9a / 64
- = 27 a / 64
- Now sequence will be a, 3a/4, 9/a16, 27a/64
- Now common difference d = a2 – a1, a3 – a2, a4 – a3
- So d = a2 – a1
- = 3a/4 – a
- = - a / 4
- So d = a3 – a2
- = 9a / 16 – 3a / 4
- = - 3a / 16
- Now a2 – a1 ≠ a3 – a2
- Therefore the sequence will not form an A.P
Reference link will be
https://brainly.in/question/8271120
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