# In which of the following situations, the sequence of numbers formed will form an A.P.?

Question:

In which of the following situations, the sequence of numbers formed will form an A.P.?

(i) The cost of digging a well for the first metre is Rs 150 and rises by Rs 20 for each succeeding metre.

(ii) The amount of air present in the cylinder when a vacuum pump removes each time 14">1414 of their remaining in the cylinder.

Q

Solution:

(i) In the given problem,

Cost of digging a well for the first meter = Rs 150

Cost of digging a well for subsequent meter is increased by Rs 20

So,
Cost of digging a well of depth one meter= Rs. 150

Cost of digging a well of depth two meters= Rs = Rs.

Cost of digging a well of depth three meters= Rs = Rs.

Cost of digging a well of depth four meters = Rs = Rs.

Thus, the costs of digging a well of different depths are

Now, for a sequence to be an A.P., the difference between adjacent terms should be equal.

Here,

$a_{1}-a=170-150$

$=20$

Also,

$a_{2}-a_{1}=190-170$

$=20$

Therefore, $a_{1}-a=a_{2}-a_{1}$

Since the terms of the sequence are at a common difference of 20, the above sequence is an A.P. with the first term as $a=150$ and common difference $d=20$.

(ii) Here, let us take the initial amount of air present in the cylinder as 100 units.

So,

Amount left after vacuum pump removes air for $1^{\text {st }}$ time $=100-\left(\frac{1}{4}\right) 100$

$=100-25$

$=75$

Amount left after vacuum pump removes air for $2^{\text {nd }}$ time $=75-\left(\frac{1}{4}\right) 75$

$=75-18.75$

$=56.25$

Amount left after vacuum pump removes air for $3^{\text {rd }}$ time $=56.25-\left(\frac{1}{4}\right) 56.25$

$=56.25-14.06$

$=42.19$

Thus, the amount left in the cylinder at various stages is $100,75,56.25,42.19, \ldots$

Now, for a sequence to be an A.P., the difference between adjacent terms should be equal.

Here,

$a_{1}-a=75-100$

$=-25$

Also,

$a_{2}-a_{1}=56.25-75$

$=-18.75$

Since, $a_{1}-a \neq a_{2}-a_{1}$

The sequence is not an A.P.

(iii)
Here, prinical (P) = 1000
Rate (r) = 10%
Amount compounded annually is given by

$A=P\left(1+\frac{r}{100}\right)^{n}$

For the first year,

$A_{1}=1000\left(1+\frac{10}{100}\right)^{1}=1100$

For the second year,

$A_{2}=1000\left(1+\frac{10}{100}\right)^{2}=1210$

For the third year,

$A_{1}=1000\left(1+\frac{10}{100}\right)^{3}=1331$

Therefore, first three terms are 1100, 1210, 1331.
The common difference between the consecutive terms are not same.
Hence, this is not in A.P.