ifa,b,c are in A.P then proof that b+c,c+a,a+b will are on A.P pkz solve any body
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a, b, c are in A.P
Then a+c = 2b
b+c, c+a, a+b
b+c+a+b
= a+c+2b
= 2b+2b [a+c = 2b]
= 4b
Again,
2(c+a)
= 2×2b [a+c = 2b]
= 4b
b+c+a+b = 2(c+a)
b+c, c+a, a+b are in A.P [Proved]
Then a+c = 2b
b+c, c+a, a+b
b+c+a+b
= a+c+2b
= 2b+2b [a+c = 2b]
= 4b
Again,
2(c+a)
= 2×2b [a+c = 2b]
= 4b
b+c+a+b = 2(c+a)
b+c, c+a, a+b are in A.P [Proved]
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