Question

The taxi fare after each km, when the fare is Rs. 15 for the first km and Rs. 8 for each additional km, does not form an AP as the total fare (in Rs.) after each km is 15, 8, 8, 8…
If true then enter 1 and if false then enter 0

Answer 1

Hint: In this question we are asked whether the given sequence of fares forms an Arithmetic progression (AP) or not. Therefore, we should first understand the definition of AP and then compare the given series with the general form of an arithmetic progression. Then, if this series fits the general form of an AP with any first term and common difference, then we can call this series to be an AP otherwise not.

Complete step by step solution:
The general form of Arithmetic Progression(AP) is given by
${{a}_{n}}={{a}_{1}}+(n-1)d.............................(1.1)$
Where ${{a}_{n}}$ is the nth term for any natural number n, ${{a}_{0}}$ is the first term and d is the common difference.
We observe that the difference between any two successive terms, ${{a}_{r+1}}$ and ${{a}_{r}}$ is given by
${{a}_{r+1}}-{{a}_{r}}=\left( {{a}_{1}}+(r+1)d \right)-\left( {{a}_{1}}+rd \right)=(r+1-r)d=d...............(1.2)$
As d does not depend on r and therefore on the position of the two successive terms in the arithmetic progression, the difference between any two successive terms in an arithmetic progression should be a constant ………………………….(1.3)
In the given sequence 15,8,8,… we note that the difference between the first two terms is $8-15=-7$ . However, the difference between the second and third terms is $8-8=0$ and similarly the difference between the next successive terms Is also zero………………………..(1.4)
However, as from (1.4) we see that the difference between the successive terms in an AP is not a constant, and thus it does not satisfy the property of an AP as listed in (1.3).
Therefore, the given sequence is not an AP.
So the statement is true so the answer is 1.

Note: We should note that we had to disprove the given statement that the sequence is in AP, therefore giving one property of AP which is not satisfied by the AP is sufficient to disprove the sequence from being an AP. However, we should note that if we had to prove that the sequence is in AP, we should have shown that its terms are in the general form as in equation (1.1) and thus it should satisfy all the properties of an AP.