# Arithmetic Progression question

• Saitama

## Homework Statement

If x ε R, the numbers 51+x+51-x, a/2, 25x
+25-x form an AP, then 'a' must lie in the interval:-
a)[1,5)
b)[2,5]
c)[5,12]
d)[12,∞)

Not required

## The Attempt at a Solution

I substituted y=5x.
The terms are in AP, so the common difference is constant. That is:-
$$\frac{a}{2}-5y-\frac{5}{y}=y^2+\frac{1}{y^2}-\frac{a}{2}$$
$$\Rightarrow a=y^2+\frac{1}{y^2}+5y+\frac{5}{y}$$

Now i am stuck here. :(

Thanks!

Think about the domain of each variable. $x\in R$ thus $y=5^x$ is in which set? And what about a?
You may want to graph the function $$a=y^2+\frac{1}{y^2}+5y+\frac{5}{y}$$ to get an idea of what the answer might be.

Think about the domain of each variable. $x\in R$ thus $y=5^x$ is in which set? And what about a?

Thanks for the reply Mentallic.

y=5x in (0,∞) set. Right?

Then a should also lie in the same set.

Can you tell me an other way except graphing? Because this question is from a test paper and i don't think graphing the function would do me any good because i suck at graphing. :D

## Homework Statement

If x ε R, the numbers 51+x+51-x, a/2, 25x
+25-x form an AP, then 'a' must lie in the interval:-
a)[1,5)
b)[2,5]
c)[5,12]
d)[12,∞)

Not required

## The Attempt at a Solution

I substituted y=5x.
The terms are in AP, so the common difference is constant. That is:-
$$\frac{a}{2}-5y-\frac{5}{y}=y^2+\frac{1}{y^2}-\frac{a}{2}$$
$$\Rightarrow a=y^2+\frac{1}{y^2}+5y+\frac{5}{y}$$

Now i am stuck here. :(

Thanks!

Now substitute 5x back in for y.
$a=5^{2 x}+(5)5^{x}+(5)5^{-x}+5^{-2 x}$​

See if you can find a minimum and/or a maximum for this.

(It's an even function.)

Now substitute 5x back in for y.
$a=5^{2 x}+(5)5^{x}+(5)5^{-x}+5^{-2 x}$​

See if you can find a minimum and/or a maximum for this.

(It's an even function.)

Hi SammyS! :)

Even function? What do you mean by that?

And i am not sure about how to find maximum and minimum of a function. I just read it once.
Do we find minimum or maximum like this: First we take the derivative of the function and then substitute it to zero.?

Last edited:
Hi SammyS! :)

Even function? What do you mean by that?

And i am not sure about how to find maximum and minimum of a function. I just read it once.
Do we minimum or maximum like this: First we take the derivative of the function and then substitute it to zero.?
If you know how to take the derivative, then do that and set it equal to zero. (This is the precalculus section.)

An even function is one whose graph is symmetric with respect to the y-axis.

You can also represent this as the sum of two hyperbolic cosine functions.

5 = eln(5), so

$\displaystyle a=5^{2 x}+5^{-2 x}+(5)5^{x}+(5)5^{-x}$
$\displaystyle =e^{2 x\,\ln5}+e^{-2 x\,\ln5}+(5)e^{x\,\ln5}+(5)e^{-x\,\ln5}$

$\displaystyle =2\cosh(2 x\,\ln5)+2(5)\cosh(x\,\ln5)$​

Now you can graph this.

If you know how to take the derivative, then do that and set it equal to zero. (This is the precalculus section.)

An even function is one whose graph is symmetric with respect to the y-axis.

You can also represent this as the sum of two hyperbolic cosine functions.

5 = eln(5), so

$\displaystyle a=5^{2 x}+5^{-2 x}+(5)5^{x}+(5)5^{-x}$
$\displaystyle =e^{2 x\,\ln5}+e^{-2 x\,\ln5}+(5)e^{x\,\ln5}+(5)e^{-x\,\ln5}$

$\displaystyle =2\cosh(2 x\,\ln5)+2(5)\cosh(x\,\ln5)$​

Now you can graph this.

Lol, i don't know about hyperbolic functions. :rofl:

Sorry, but i did not know that we will have to use Calculus in this question, so i posted it in precalculus section. :)

Ok i go like this:-
I find the derivative of the given function which comes out to be:-
$$25^xlog_e25-25^{-x}log_e25+5(5^xlog_e5)-5(5^{-x}log_e5)$$

When i substitute it to zero, i get stuck.. Here are the steps:-
$$log_e25(25^x-25^{-x})=-5log_e5(5^x-5^{-x})$$
$$\frac{25^x-25^{-x}}{5^x-5^{-x}}=\frac{-5}{2}$$

What should i do next?

Thanks for the reply Mentallic.

y=5x in (0,∞) set. Right?

Then a should also lie in the same set.
Yes, that's right

Can you tell me an other way except graphing? Because this question is from a test paper and i don't think graphing the function would do me any good because i suck at graphing. :D

Oh, yep ofcourse. Graphing calculators are just the cheap and brainless way of figuring out the answer anyway

Looking at the function $$a=5^{2x}+5^{-2x}+5^{1+x}+5^{1-x}$$

finding a turning point for this function we set the derivative equal to zero. In short, by using the rational root theorem, you should find $5^x=\pm 1$ but 5x>0 so 5x=1 is the only valid solution, thus x=0 is where the turning point occurs. You can go further and show this is a minimum turning point, so at x=0, $a=5^0+5^0+5^{1+0}+5^{1+0}=12$ and clearly a is always increasing since all the powers are greater than zero, and 5x approaches infinite as x approaches infinite, so we have the answer.

I find the derivative of the given function which comes out to be:-
$$25^xlog_e25-25^{-x}log_e25+5(5^xlog_e5)-5(5^{-x}log_e5)$$
Yes good work, that's correct.

When i substitute it to zero, i get stuck..
Because you shouldn't be plugging in x=0, that will give you the gradient of the original function at x=0, what you instead want to do is solve the derivative equal to 0. Multiply through by 25x and to make things simpler, let y=5x. Then use the rational root theorem to find the roots.

Yes, that's right

Oh, yep ofcourse. Graphing calculators are just the cheap and brainless way of figuring out the answer anyway

We aren't allowed calculators during an exam. :rofl:

Yes good work, that's correct.

Because you shouldn't be plugging in x=0, that will give you the gradient of the original function at x=0, what you instead want to do is solve the derivative equal to 0. Multiply through by 25x and to make things simpler, let y=5x. Then use the rational root theorem to find the roots.

I did not plugged x=0, i solved the derivative equal to 0.
Sorry for my confusing English.

Here i go again:-
let y=5x
then continuing my above equation,
$$\frac{y^2-y^{-2}}{y-y^{-1}}=\frac{-5}{2}$$
Solving i get:-
$$\frac{y^2+1}{y}=\frac{-5}{2}$$
$$\Rightarrow 2y^2+5y+2=0$$

Solving i get, y=-2 and -1/2.
that is, 5^x=-2 and -1/2.
This is funny, we need to have y>0. :D

I don't understand where i am wrong.

Bump :)

I did not plugged x=0, i solved the derivative equal to 0.
Sorry for my confusing English.
Ahh ok, then your problem lies elsewhere :tongue:

Here i go again:-
let y=5x
then continuing my above equation,
$$\frac{y^2-y^{-2}}{y-y^{-1}}=\frac{-5}{2}$$
Solving i get:-
$$\frac{y^2+1}{y}=\frac{-5}{2}$$
How did you get from the first equation to the next? I suggest you get rid of the fractions, so that involves multiplying through by y2. You'll have a polynomial of degree 4 to solve, which is why I mentioned the rational root theorem.

$$\Rightarrow a=y^2+\frac{1}{y^2}+5y+\frac{5}{y}$$

You can see at once that a can go to infinity.

To find the minimum of a, find the derivative with respect to y. You did it with respect to x, but it differs only by a factor of 5xln(5)

You get the equation

$$2y-\frac{2}{y^3}+5-\frac{5}{y^2}=0$$

factor out 5 and 2y:

$$5(1-\frac{1}{y^2})+2y(1-\frac{1}{y^4})=0$$

Notice that $$1-\frac{1}{y^4}=(1-\frac{1}{y^2})(1+\frac{1}{y^2})$$

$$5(1-\frac{1}{y^2})+2y(1-\frac{1}{y^2})(1+\frac{1}{y^2})=0$$

$$(1-\frac{1}{y^2})(5+2y(1+\frac{1}{y^2}))=0$$
One solution is y2-1=, y=1. The other solution of
$$5+2y(1+\frac{1}{y^2})=0$$
leads to negative roots, which are inappropriate. So the only extremum is at y=1, and it is minimum.

ehild

Ahh ok, then your problem lies elsewhere :tongue:

How did you get from the first equation to the next? I suggest you get rid of the fractions, so that involves multiplying through by y2. You'll have a polynomial of degree 4 to solve, which is why I mentioned the rational root theorem.

Lol, i have never used that theorem. :D
I think i will have to refer wiki.

You can see at once that a can go to infinity.

To find the minimum of a, find the derivative with respect to y. You did it with respect to x, but it differs only by a factor of 5xln(5)

You get the equation

$$2y-\frac{2}{y^3}+5-\frac{5}{y^2}=0$$

factor out 5 and 2y:

$$5(1-\frac{1}{y^2})+2y(1-\frac{1}{y^4})=0$$

Notice that $$1-\frac{1}{y^4}=(1-\frac{1}{y^2})(1+\frac{1}{y^2})$$

$$5(1-\frac{1}{y^2})+2y(1-\frac{1}{y^2})(1+\frac{1}{y^2})=0$$

$$(1-\frac{1}{y^2})(5+2y(1+\frac{1}{y^2}))=0$$
One solution is y2-1=, y=1. The other solution of
$$5+2y(1+\frac{1}{y^2})=0$$
leads to negative roots, which are inappropriate. So the only extremum is at y=1, and it is minimum.

ehild

Thank you ehild. I get it now.
I was getting only negative roots, now my problem is solved, thank you.

You are welcome.
Advice: never ever simplify with something which can be zero.

ehild

You are welcome.
Advice: never ever simplify with something which can be zero.

ehild

Thanks for the advice.
I will remember that. :)