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Question
In the following situation, involved make an arithmetic progression? and why?
The taxi fare after each km when the fare is ₹ 15 for the first km and ₹ 8 for each additional km.
Solution
It can be observed that
Taxi fare for 1st km = 15
Taxi fare for first 2 km = 15 + 8 = 23
Taxi fare for first 3 km = 23 + 8 = 31
Taxi fare for first 4 km = 31 + 8 = 39
a2 - a1 = 23 - 15 = 8
a3 - a2 = 31 - 23 = 8
a4 - a3 = 39 - 31 = 8
Clearly 15, 23, 31, 39 … forms an A.P. because every term is 8 more than the preceding term.
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