Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • In which of the following situations, does the list of numbers involved make an arithmetic progression, and why? (ii) The amount of air present in a cylinder when a vacuum pump removes 1 by 4 of the air remaining in the cylinder at a time.

Q : 1     In which of the following situations, does the list of numbers involved make an arithmetic  progression, and why?

              (ii) The amount of air present in a cylinder when a vacum pump removes   \small \frac{1}{4}  of the air  remaining in the cylinder at a time.

Answers (1)

best_answer

It is given that 
vacum pump removes   \small \frac{1}{4}  of the air  remaining in the cylinder at a time
Let us take initial quantity of air = 1

Now, quantity of air removed in first step = 1/4

Remaining quantity after 1st step

= 1-\frac{1}{4}= \frac{3}{4}

Similarly, Quantity removed after 2nd step = Quantity removed in first step \times  Remaining quantity after 1st step

=\frac{3}{4}\times \frac{1}{4}= \frac{3}{16}
Now,

Remaining quantity after 2nd step would be = Remaining quantity after 1st step - Quantity removed after 2nd step

=\frac{3}{4}- \frac{3}{16}= \frac{12-3}{16}= \frac{9}{16}
Now, we can clearly see that

After second step the difference between second and first and first and initial step is not the same, hence

the common difference (d)  is not same after every step 

Therefore, it is not an AP

Posted by

Gautam harsolia

View full answer

Similar Questions

Trending Articles/News

anam-khan

Board Exam Date 2025 Live: Bihar Class 10, 12 model papers out, time table soon; updates on CBSE, UP

Latest Question