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Chapter 8: Problem 678

If \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are in G. P., \(\mathrm{a}, \mathrm{x}, \mathrm{b}\) are in \(\mathrm{A}\). P. and \(\mathrm{b}, \mathrm{y}, \mathrm{c}\) are in A. P., then \((\mathrm{a} / \mathrm{x})+(\mathrm{c} / \mathrm{y})=\) (A) 1 (B) \((1 / 2)\) (C) 2 (D) 4

Short Answer

Expert verified
The short answer for the given question is: \(\boxed{2}\) (C).

Step by step solution

01

Write the relationships for G.P. and A.P.

Since a, b, and c are in G.P., we can write the relationship as: \[b = ar\] and \[c = ar^2\], where r is the common ratio. Since a, x, and b are in A.P., we can write this relationship as: \[x = (a+b)/2\] Similarly, since b, y, and c are in A.P., we can write the relationship as: \[y = (b+c)/2\]
02

Substitute the relationships into the given expression

Using the relationships we found, we can now substitute them into the expression \((a/x) + (c/y)\): \[\frac{a}{(a+b)/2} + \frac{c}{(b+c)/2}\]
03

Simplify the expression

Now, we simplify the expression by multiplying by the denominators so that our term will have no fractions: \[\frac{2a}{a+b} + \frac{2c}{b+c}\]
04

Substitute b using G.P. relationship

Substitute b = ar in the expression: \[\frac{2a}{a+ar} + \frac{2c}{ar+ar^2}\]
05

Simplify further by factoring out common terms

Factor out common terms in the denominators: \[\frac{2a}{a(1+r)} + \frac{2c}{ar(1+r)}\]
06

Combine fractions with a common denominator

Combine the fractions by finding the common denominator, which in this case is a(1+r): \[\frac{2a+2c}{a(1+r)}\]
07

Substitute c using G.P. relationship

Now we can substitute c = ar^2 in the expression: \[\frac{2a+2ar^2}{a(1+r)}\]
08

Simplify the expression further

Factor out a common term in the numerator and simplify: \[\frac{2a(1+r^2)}{a(1+r)}\]
09

Cancel out the common terms and find the result

Cancel out the common terms a(1+r) in the numerator and denominator: \[\frac{2(1+r^2)}{1+r} = \frac{2(1+r)}{1+r}\] Finally, by canceling out (1+r) in the numerator and denominator, we get the solution: \[1+1 = \boxed{2}\] #Answer#(C) 2

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