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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 $\frac{1}{4}$ of their remaining in the cylinder.

Q

(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 $\frac{1}{4}$ of their remaining in the cylinder.

Q

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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,

Also,

Therefore,

Since the terms of the sequence are at a common difference of 20, the above sequence is an A.P. with the first term asand common difference.

(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^{st} time=

Amount left after vacuum pump removes air for 2^{nd} time=

Amount left after vacuum pump removes air for 3^{rd} time=

Thus, the amount left in the cylinder at various stages is

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

Here,

Also,

Since,

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.

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