GP & the Interval

Algebra Level 3

If an infinite GP of real numbers has second term $x$ and sum $4,$ where does $x$ belong?

(-8, 1]

(0, 2)

[1, 8)

[-4, 1]

4 solutions

Sanjeet Raria
Oct 8, 2014

Let $r$ be the common ratio. $Sum=\frac{x/r}{1-r}=4$ $x=4(r-r^2)$

The sum of infinite GP converges $\Rightarrow r\in (-1,1)$ $\Rightarrow r-r^2\in(-2,\frac{1}{4})$ $\Rightarrow x\in(-8,1]$

Note: You should be careful about the $r = 0$ case.

Calvin Lin Staff - 6 years, 8 months ago

Even if r=0, what difference would it make as far as the question is concerned? In that case, first term can be 4& x=0.

Sanjeet Raria - 6 years, 8 months ago

Yes it can.

Note that if $r = 0$ , then $x/r$ is undefined. Hence, implicit in your solution, is the assumption that $r \neq 0$ . Thus, we have to deal with this case seperately.

Calvin Lin Staff - 6 years, 8 months ago

@Calvin Lin – Okay sir, i would write in my solution further that $r\ne0$ .

But then in a separate note can not i right that if $x=0, \Rightarrow r=0$ for the sake of our question?? Therefore i would not violate my solution & will even cover the case of $r=0$ ??

Sanjeet Raria - 6 years, 8 months ago

@Sanjeet Raria – This is a ques asked in IIT-JEE previous year. :)

Sandeep Bhardwaj - 6 years, 8 months ago

Since in case of G.P. , none of its terms can be 0. And also common ratio $r$ can not be 0. In this case, I would like to add that $x$ belongs to $(-8,1]$ does necessarily mean that all the real numbers between $(-8,1]$ will fulfill the required conditions, But it will be more accurate to say that whatever the values of $x$ will satisfy the given conditions, those values will lie in the interval $(-8,1]$

Sandeep Bhardwaj - 6 years, 8 months ago

Can you please explain how r-r^2 belong to (-2, (1/4)) when r belong to (-1,1) ?

Gautham Shankar - 4 years, 2 months ago

Think about the maximum and minimum value for $r-r^{2}$ when you know $r \in (-1,1)$

You can use differentiation too.

Paras Lehana - 3 years, 8 months ago

Can you please explain how r-r^2 belong to (-2, (1/4)) when r belong to (-1,1) ?

Gautham Shankar - 3 years, 10 months ago

probably ... r-r^2= -(r-0.5)^2+0.25 so the maximum value of this expression is 0.25 . and if you plug in r=-1 you will get -2

Muhammad Salem - 3 years, 4 months ago

A Former Brilliant Member
Nov 8, 2017

As r < 1 , we know it has to be limited between

$S_n = 4$

$\frac{a}{1-r} = 4$

x = 4r (1-r)

You can now plug in the limiting values of r to get the answer.

$x\in(-8,1]$

In equation x = 4r(1-r) if we plug r = -1 we get x =-8 and for r= 1 we get x = 0. Then why right side of interval is 1?

Sachin Patil - 2 years, 7 months ago

You can try to derivate x=4r(1-r), you will find that the maximum value of 4r(1-r) occours when r=1/2, so x=1. Rembember that 4r -4r^2 is a parabola with down concavity, so the maximum valuem is in the mean of roots (0;1)

Felipe Mello - 1 year, 2 months ago

Shubhrajit Sadhukhan
Oct 2, 2020

Let the initial term of the GP be $a'$ and the common ratio be $r$ .

$S_∞ = 4$

$\dfrac{a'}{1 - r} = 4$

$\dfrac{a'r}{1 - r} = 4r$

$\dfrac{x}{1 - r} = 4r$

$x = -4r^2 + 4r$

When $r = -1$ , $x = -8$

When $r = 1$ , $x = 0$

Now, $x$ (a function of $r$ ) is a downward parabola since $a (= -4)$ is negative. So, maximum value of $x$ occurs at $r = \frac{-b}{2a} = \frac{1}{2}$ which is between $-1$ and $1$ . (See plot here )

When $r = \dfrac{1}{2}$ , $x = 1$

So, $\boxed{x \in (-8, 1]}$

Hakan Eskici
Apr 23, 2019

Where did I make a mistake? could anyone help me?

See my solution

Shubhrajit Sadhukhan - 8 months, 2 weeks ago

GP & the Interval

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