Enough manipulations?

Algebra Level 3

If a , b , c a,b,c and d d follows a harmonic progression (in that order), which of these answer choices equals to a b + b c + c d ab + bc + cd ?

3 a c 3ac ( a + b ) ( c + d ) (a+b)(c+d) None of these choices 3 a d 3ad

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.

2 solutions

Arjen Vreugdenhil
Sep 18, 2015

A more elementary solution: a = 1 n ; b = 1 n + m ; c = 1 n + 2 m ; d = 1 n + 3 m . a = \frac{1}{n};\;b=\frac{1}{n+m};\;c=\frac{1}{n+2m};\;d=\frac{1}{n+3m}. The given sum is 1 n ( n + m ) + 1 ( n + m ) ( n + 2 m ) + 1 ( n + 2 m ) ( n + 3 m ) \frac{1}{n(n+m)}+\frac{1}{(n+m)(n+2m)}+\frac{1}{(n+2m)(n+3m)} = ( n + 2 m ) ( n + 3 m ) + n ( n + 3 m ) + n ( n + m ) n ( n + m ) ( n + 2 m ) ( n + 3 m ) =\frac{(n+2m)(n+3m)+n(n+3m)+n(n+m)}{n(n+m)(n+2m)(n+3m)} = 3 n 2 + 9 n m + 6 m 2 n ( n + m ) ( n + 2 m ) ( n + 3 m ) = 3 ( n + m ) ( n + 2 m ) n ( n + m ) ( n + 2 m ) ( n + 3 m ) =\frac{3n^2+9nm+6m^2}{n(n+m)(n+2m)(n+3m)}=\frac{3(n+m)(n+2m)}{n(n+m)(n+2m)(n+3m)} = 3 n ( n + 3 m ) = 3 a d . =\frac{3}{n(n+3m)}=3ad.

Abhinav Whizkid
Aug 28, 2015

we know that if A1,A2,A3......An are in HP then : A1A2+A2A3+A3A4+.......+An-1An=(n-1)A1An

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...