The fare of taxi is #15 for the first km and rises by #10 for each additional km. weather this situation involved in Arithmetic progression ? why?
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Answer:
Here it is
Step-by-step explanation:
The taxi fare for first km=15
The taxi fare for 2nd km=15+8=23
The taxi fare for 3rd km=15+8+8=31 and so on,
Therefore, the taxi fare form the sequence 15,23,31,...
Clearly, it is an AP with first term a=15 and common difference d=8
Therefore, answer is 0.
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