The taxi fare after each km,when the fare is | vedclass

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(B) No,the statement is false. The total fare (in $Rs.$) after each $km$ is calculated as follows:For $1$ $km$: $15$For $2$ $km$: $15 + 8 = 23$For $3$

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Vedclass · Answer

(B) No,the statement is false. The total fare (in $Rs.$) after each $km$ is calculated as follows:
For $1$ $km$: $15$
For $2$ $km$: $15 + 8 = 23$
For $3$ $km$: $15 + 2 \times 8 = 31$
For $4$ $km$: $15 + 3 \times 8 = 39$
Thus,the sequence of total fares is $15, 23, 31, 39, \ldots$.
Let $t_1 = 15, t_2 = 23, t_3 = 31, t_4 = 39$.
Now,calculating the common difference:
$t_2 - t_1 = 23 - 15 = 8$
$t_3 - t_2 = 31 - 23 = 8$
$t_4 - t_3 = 39 - 31 = 8$
Since the difference between consecutive terms is constant (common difference $d = 8$),the sequence forms an Arithmetic Progression $(AP)$.

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 $15, 8, 8, 8, \ldots$. Is the statement true? Give reasons.

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