The third proportional of the following numbers (x - y)2, (x2 - y2) = ?
A. (x + y3) (x + y4)
B. (x + y)4 (x - y)2
C. (x - y) (x + y)2
D. (x + y)2 (x - y)3
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & {\text{Let,}} \cr & a = {\left( {x - y} \right)^2}, \cr & b = \left( {{x^2} - {y^2}} \right){\text{and}} \cr & c\,\,{\text{be}}\,{\text{the}}\,{\text{third}}\,{\text{proportional}} \cr & {\text{Therefore}}\,\,\,a:b::b:c \cr & i.e.\,\,\,c = \frac{{{b^2}}}{a} \cr & \Rightarrow c = \frac{{{{\left( {{x^2} - {y^2}} \right)}^2}}}{{\left( {x - y} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{{{{\left( {x - y} \right)}^2}{{\left( {x + y} \right)}^2}}}{{\left( {x - y} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \left( {x - y} \right){\left( {x + y} \right)^2} \cr} $$This Question Belongs to Arithmetic Ability >> Algebra
Join The Discussion
Comments ( 2 )
Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
A. $$\frac{{20}}{{27}}$$
B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$
This formula is not for third proportion .
I think this question is wrong because the solution is not for this question please verify question with answer and make possible corrections
Thanks in advance