Given that (1/b-2a),(1/b),(1/b-2c) are in arithmetic progression, prove that a(b-c),ac,c(b-a) are also in AP.?

Answer 1

If this is true, then 2/b = 1/(b- 2a) +1/(b - 2c) = (2b - 2a - 2c)/(b -2a)(b - 2c). Then, (b - 2a)(b - 2c) =b( b - a - c). Expanding the products, we get ,

b^2 - 2(a + c)b + 4ac = b^2 - b(a + c), so that

4ac = b(a + c) => b = 4ac/(a + c), supposing a +c <>0

In addition, a(b - c) + c(b - a) = ab - ac + bc -ac = b(a + c) = 4ac . This proves a(b- c), 2ac and c(b - a) are in arithmetic progression, but not as stated

Answer 2

I'll have to assume that the original sequence is
1/(b-2a), 1/b, 1/(b-2c), and not (1/b) - 2a, 1/b, (1/b) - 2c, because I don't get a conclusive answer the second way.

If these are three values in an arithmetic sequence, then you add the same number in one term to get the next. This means the differences in consecutive terms should be the same, so:
1/b - 1/(b-2a) = 1/(b-2c) - 1/b
2/b - 1/(b-2a) = 1/(b-2c)
2 - b/(b-2a) = b/(b-2c)
2(b-2c) - b(b-2c)/(b-2a) = b
2(b-2c)(b-2a) - b(b-2c) = b(b-2a)
2(b-2c)(b-2a) - b^2 + 2bc = b^2 - 2ab
2(b-2c)(b-2a) + 2bc = 2b^2 - 2ab
(b-2c)(b-2a) + bc = b^2 - ab
(b-2c)(b-2a) = b^2 - ab - bc
b^2 - 2bc - 2ab + 4ac = b^2 - ab - bc
-2bc - 2ab + 4ac = -ab - bc
-bc - ab + 4ac = 0
4ac = ab + bc

Likewise, if we assume a(b-c), ac, c(b-a) is part of an arithmetic sequence, then this can only happen when the differences in terms are the same, so we have:
c(b-a) - ac = ac - a(b-c)
c(b-a) = 2ac - a(b-c)
bc - ac = 2ac - ab + ac
bc = 4ac - ab
ab + bc = 4ac

We get the same result as last time, which proves that if the first sequence is arithmetic, then the second one is too. If we wanted to do this proof more fomally, we could leave the first half alone, then in the second half take ab + bc = 4ac and work backwards until we arrive at c(b-a) - ac = ac - a(b-c).

Answer 3

in the event that they are an AP, then the transformations are equivalent. as a result: a million/b – a million/a = a million/c – a million/b Multiply via by skill of abc and get ac – bc = ab – ac yet as a results of fact of this bc, ac, ab sort an AP as a results of fact consecutive transformations are equivalent. QED