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Let the terms be (b+c-a), (c+a-b) nd (a+b-c)
Second term- first term=c+a-b-(b+c-a)= 2a-2b= 2(a-b)(eq1)
Third term-second term=(a+b-c)-(c+a-b)= 2b-2c= 2(b-c)(eq2)
From the question, a,b,c are in AP
So b-a=c-b
Multiplying by -1 on both sides
a-b=b-c
Multiplying 2 on both sides
2(a-b)=2(b-c)
Therefore eq1=eq2
Since the common difference is same,
therefore (b+c-a), (c+a-b) nd (a+b-c) are in AP
U can ask any doubts...Just comment below
Hope it helps u☺️☺️☺️☺️❤️
Second term- first term=c+a-b-(b+c-a)= 2a-2b= 2(a-b)(eq1)
Third term-second term=(a+b-c)-(c+a-b)= 2b-2c= 2(b-c)(eq2)
From the question, a,b,c are in AP
So b-a=c-b
Multiplying by -1 on both sides
a-b=b-c
Multiplying 2 on both sides
2(a-b)=2(b-c)
Therefore eq1=eq2
Since the common difference is same,
therefore (b+c-a), (c+a-b) nd (a+b-c) are in AP
U can ask any doubts...Just comment below
Hope it helps u☺️☺️☺️☺️❤️
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