Sequence or not?

Algebra Level pending

9 ( 25 a 2 + b 2 ) + 25 ( c 2 − 3 c a ) = 15 b ( 3 a + c ) \large 9(25a^2+b^2) + 25(c^2-3ca) = 15b(3a+c)

a , b , c a,b,c (in some order) are in which sequence?

Note :- a , b , c a,b,c are real numbers.

Details :- a , b , c a,b,c (in that order) are

• in AP if 2 b = a + c 2b=a+c

• in GP if b 2 = a c b^2=ac

• in HP if b = 2 a c a + c b=\frac {2ac}{a+c}

None of the above Geometric Progression (GP) Harmonic Progression (HP) Arithmetic Progression (AP)

Parth Sankhe by Parth Sankhe , 27, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Mark Hennings
Dec 4, 2018

If we put A = 15 a A=15a , B = 3 b B=3b and C = 5 c C=5c , this equation becomes A 2 + B 2 + C 2 − A C = A B + B C A 2 + B 2 + C 2 − A B − A C − B C = 0 1 2 [ ( A − B ) 2 + ( A − C ) 2 + ( B − C ) 2 ] = 0 \begin{aligned} A^2 + B^2 + C^2 - AC & = \; AB + BC \\ A^2 + B^2 + C^2 - AB - AC - BC & = \; 0 \\ \tfrac12\big[(A-B)^2 + (A-C)^2 + (B-C)^2\big] &= \; 0 \end{aligned} so that A = B = C A = B = C . But then 1 2 ( A + 5 B ) = 3 C \tfrac12(A + 5B) = 3C , and so 1 2 ( a + b ) = c \tfrac12(a+b) = c so a , c , b a,c,b (in that order) are in A P \boxed{\mathrm{AP}} .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...