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Q1(i):
In which of the following situations, does the list of numbers involved make an arithmetic progression, and why? (i) The taxi fare after each km when the fare is ₹ 15 for the first km and ₹ 8 for each additional km.

Solution :

Given:

  • Fare for the first kilometer ($a_1$) = ₹ $15$.
  • Additional fare for every subsequent kilometer ($d$) = ₹ $8$.

To Find:

Determine whether the sequence of taxi fares for $1, 2, 3, 4, \dots$ kilometers forms an Arithmetic Progression (AP) and provide the justification.

Definition:

An Arithmetic Progression is a sequence of numbers in which the difference between any two consecutive terms is constant. This constant difference is called the common difference ($d$). If the terms are $a_1, a_2, a_3, \dots, a_n$, then the sequence is an AP if $(a_2 - a_1) = (a_3 - a_2) = (a_4 - a_3) = d$.

Step 1: Calculating the terms of the sequence

Let $a_n$ be the fare for $n$ kilometers.

  • Fare for 1st km ($a_1$) = $15$
  • Fare for 2nd km ($a_2$) = $a_1 + 8 = 15 + 8 = 23$
  • Fare for 3rd km ($a_3$) = $a_2 + 8 = 23 + 8 = 31$
  • Fare for 4th km ($a_4$) = $a_3 + 8 = 31 + 8 = 39$

The sequence of fares is: $15, 23, 31, 39, \dots$

Step 2: Verifying the common difference

To check if the sequence is an AP, we calculate the difference between consecutive terms:

  • Difference 1 ($d_1$) = $a_2 - a_1 = 23 - 15 = 8$
  • Difference 2 ($d_2$) = $a_3 - a_2 = 31 - 23 = 8$
  • Difference 3 ($d_3$) = $a_4 - a_3 = 39 - 31 = 8$

Step 3: Conclusion

Since the difference between consecutive terms is constant ($d = 8$), the sequence satisfies the definition of an Arithmetic Progression.

Final Answer: Yes, the situation forms an Arithmetic Progression because the fare increases by a constant amount of ₹ 8 for each additional kilometer, resulting in a common difference of 8.


More Questions from Class 10 Mathematics Arithmetic Progression EXERCISE 5.1


CBSE Solutions for Class 10 Mathematics Arithmetic Progression


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