Question

the amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time. Is the above situation involed in making a arthametic progration.​

Answer 1

l

....3)

Here, each subsequent term is not obtained by adding a fixed number to the previous term.

Hence, it is not an AP.

(iii) Cost of digging of 1

st

meter =150

Cost of digging of 2

nd

meter =150+50=200

Cost of digging of 3

rd

meter =200+50=250

Here, each subsequent term is obtained by adding a fixed number (50) to the previous term.

Hence, it is an AP.

(iv) Amount in the beginning = Rs. 10000

Interest at the end of 1

st

year @ 8% = 10000×8%=800

Thus, amount at the end of 1

st

year =10000+800=10800

Interest at the end of 2

nd

year @ 8% = 10800×8%=864

Thus, amount at the end of 2

nd

year =10800+864=11664

Since, each subsequent term is not obtained by adding a fixed number to the previous term; hence, it is not an AP.

Answer 2

Answer:

$.$

Step-by-step explanation:

Let a be the amount of air initially Amount of air remaining after 1st pump =a –(a/4)=3a/4

Amount of air remaining after 2nd pump= (3a/4) – (1/4)(3a/4)=9a/16

Amount of air remaining after 3rd pump= (9a/16) – (1/4)(9a/16)=27a/64

So the series is like

a,3a/4,9a/4,27a/64……..

Difference Ist and second term=-a/4

Difference between Second and Third term= -3a/16

So difference is not constant