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Asked by Rksanthosh Last Modified
Certainly! As an experienced tutor registered on UrbanPro, I'd like to guide you through this problem using the properties of an arithmetic progression (AP).
Firstly, it's important to understand that in an arithmetic progression, each term is obtained by adding a fixed number (called the common difference) to the previous term.
Given: The arithmetic mean (AM) between the ath and bth terms of an AP is equal to the AM between the cth and dth terms.
Let's denote:
Now, we know that the arithmetic mean (AM) between two terms is the average of those terms.
So, the AM between the ath and bth terms is: a+(a+(b−1)d)2=2a+bd−d22a+(a+(b−1)d)=22a+bd−d
Similarly, the AM between the cth and dth terms is: a+(a+(c−1)d)2=2a+cd−d22a+(a+(c−1)d)=22a+cd−d
Given that these two AMs are equal, we have: 2a+bd−d2=2a+cd−d222a+bd−d=22a+cd−d
Now, let's simplify this equation: 2a+bd−d=2a+cd−d2a+bd−d=2a+cd−d bd=cdbd=cd
Now, subtracting 'a' from both sides: b−a=c−ab−a=c−a
Adding 'a' to both sides: a+b=a+ca+b=a+c
Finally, subtracting 'a' from both sides again: a+b=c+da+b=c+d
Thus, we have proven that if the arithmetic mean (AM) between the ath and bth terms of an arithmetic progression (AP) is equal to the AM between the cth and dth term, then a+b=c+da+b=c+d.
This demonstrates the fundamental property of arithmetic progressions, and understanding it can aid in solving various problems in mathematics. If you need further clarification or assistance with similar problems, UrbanPro is an excellent platform to find skilled tutors who can provide personalized guidance.
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