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Chapter 8: Problem 666

If \([1 /(\mathrm{b}-\mathrm{c})],[1 /(2 \mathrm{~b}-\mathrm{x})]\) and \([1 /(\mathrm{b}-\mathrm{a})]\) are in A. P., then \(\mathrm{a}-(\mathrm{x} / 2), \mathrm{b}-(\mathrm{x} / 2), \mathrm{c}-(\mathrm{x} / 2)\) are in (A) A. P. (B) G. P. (C) H. P. (D) A. G P.

Short Answer

Expert verified
The expressions \(a - \frac{x}{2}\), \(b - \frac{x}{2}\), and \(c - \frac{x}{2}\) are in Harmonic Progression (H.P.). So, the correct answer is (C) H.P.

Step by step solution

01

Apply A.P. condition to the given terms

Since the given three terms are in A.P., it means their consecutive differences are equal. Thus, we have: \(\frac{1}{2b-x} - \frac{1}{b-c} = \frac{1}{b-a} - \frac{1}{2b-x}\)
02

Simplify the equation and solve for x

We will simplify the equation by cross multiplying: \(\frac{2b-x}{(b-c)(2b-x)} = \frac{b-a}{(b-c)(b-a)}\) Now, since the fractions are equal, the numerators can be set equal: \(2b - x = (b-a)\) Solve for x: \(x = 2b - (b-a) = b + a\)
03

Replace given x in the expressions

Now that we have the value of x, we will replace it into the expressions we need to check for their progression: \(a - \frac{x}{2} = a - \frac{a+b}{2}\) \(b - \frac{x}{2} = b - \frac{a+b}{2}\) \(c - \frac{x}{2} = c - \frac{a+b}{2}\)
04

Check for progression

To check if these terms are in A.P., we look at their common differences: \((b - \frac{a+b}{2}) - (a - \frac{a+b}{2}) = b - a\) \((c - \frac{a+b}{2}) - (b - \frac{a+b}{2}) = c - b\) Since \(b - a \neq c - b\), these terms are not in A.P. To check for G.P., we look at their ratios: \(\frac{b - \frac{a+b}{2}}{a - \frac{a+b}{2}} \quad \text{and} \quad \frac{c - \frac{a+b}{2}}{b - \frac{a+b}{2}}\) However, since the terms do not share a common ratio, they are not in a G.P. Since the given terms are in A.P., their reciprocals are in H.P. by definition: \(\frac{1}{a - \frac{x}{2}}, \frac{1}{b - \frac{x}{2}}, \frac{1}{c - \frac{x}{2}}\) Therefore, our answer is (C) H.P.

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Most popular questions from this chapter

If \([(b+c-a) / a],[(c+a-b) / b],[(a+b-c) / c]\) are in A. \(P\) then \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are in (A) G P. (B) A. P. (C) H. P. (D) A. G P.

Sum to infinity of the series \(1+(4 / 5)+\left(7 / 5^{2}\right)+\left(10 / 5^{3}\right)+\ldots\) is (A) \((5 / 16)\) (B) \((35 / 16)\) (C) \((16 / 35)\) (D) \((7 / 16)\)

\([1 /(2 \times 5)]+[1 /(5 \times 8)]+[1 /(8 \times 11)]+\ldots 100\) terms (A) \([(25) /(160)]\) (B) \((1 / 6)\) (C) \([(25) /(151)]\) (D) \([(25) /(152)]\)

If \((1 / a),(1 / H),(1 / b)\) are in A. P. then \([(H+a) /(H-a)]\) \(+[(\mathrm{H}+\mathrm{b}) /(\mathrm{H}-\mathrm{b})]=\) (A) 2 (B) 4 (C) 0 (D) 1

The series \(1.1 !+2.2 !+3.3 !+\ldots+\) n. \(n !=\) \((\mathrm{A})(\mathrm{n}+1) !-\mathrm{n}\) (B) \((n+1) !-1\) (C) \(n !-1+n\) (D) \(n !+1-n\)
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