GP and HP at the Same Time!

Algebra Level 3

Given that $a,b,c$ are in a geometric progression; $a-b,c-a,b-c$ are in a harmonic progression.

What is the value of $a+4b+c$ ?

You can try my other Sequences And Series problems by clicking here : Part II and here : Part I.

The answer is 0.

2 solutions

Sanjeet Raria
Oct 9, 2014

Nice question,

$a,b,c$ are in GP.

Let $\large a=x, b=xr,c=xr^2$ $x\neq 0, r\neq 1$ $\frac{1}{a-b},\frac{1}{c-a},\frac{1}{b-c}\text {are in HP}$ $\large\Rightarrow \frac{2}{c-a}=\frac{1}{a-b}+\frac{1}{b-c}$ $\Rightarrow \frac{2}{x(r^2-1)}=\frac{1}{x(1-r)}+\frac{1}{xr(1-r)}$ Multiplying throughout by $x(1-r)$ $\large\Rightarrow \frac{-2}{1+r}=1+\frac{1}{r}$ $\Rightarrow r^2+4r+1=0$ $\large\Rightarrow x(r^2+4r+1)=0$ $\Rightarrow x+4xr+xr^2=0$ $\Rightarrow a+4b+c=\boxed0$

Note that $a-b,c-a,b-c$ are in H.P. implies that $\frac1{a-b},\frac1{c-a},\frac1{b-c}$ are in $\large\fbox{AP}$ instead.

Kenny Lau - 6 years, 8 months ago

Since h.p so,

$c - a = \dfrac{2(a - b)(b - c)}{a - c}$

$(a - c)^2 = 2( b -a)(b - c)$

$(a - c)^2 = 2(b^2 - b(a + c) + ac)$

$(a -c)^2 = 4ac - 4b(a + c)$

$(a - c)^2 + 4ac = 8ac - 4(a + c)$

$(a + c)^2 - 8ac = -4(a +c)$

$( a+ c)^2 - 16ac = -2b(a + 4b + c)$

$( a + 4b + c)( a - 4b + c) = -2b(a + 4b + c)$

$( a + 4b + c)( a - 2b + c) = 0$

$a + 4b + c =0$

This will only be the solution , instead many of us would directly cancel off (a + 4b + c) from LHS and RHS

$a - 2b + c \neq 0$ because if it is equal to zero then $a + c = 2b$ which implies that a , b , c are in A.P but it is given that a , b ,c are in G.P.

U Z - 6 years, 4 months ago

Ivan Koswara
Feb 12, 2015

Stupid solution:

Note that the condition is invariant under scalar multiplication (if $a,b,c$ satisfies the condition, then so must be $ka,kb,kc$ for any $k$ ). Thus the sum must also be invariant under scalar multiplication (otherwise all real numbers are possible as the answer); the only possible answer is thus $\boxed{0}$ .

A non-stupid solution (that doesn't abuse the uniqueness of the answer) is given already.

GP and HP at the Same Time!

You can try my other Sequences And Series problems by clicking here : Part II and here : Part I.

The answer is 0.

2 solutions

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