In AP three consecutive terms are 5^1+X + 5^1-X, a/2, 25^x+25^-X then find a=?

Hi Halak,

As they are in AP, we have

a = 5^1+x + 5^1-x + 25^x + 25^-x.

So say 5^x = t, then,

a = 5t + 5/t + t² + 1/t².

Now 5t+5/t ≥ 10 --------- (AM-GM Inequality)

and t²+1/t² ≥ 2 --------- (AM-GM Inequality)

So a ≥ 12, are the values that "a" can take.

Hope that helps.

Best Regards,

Ashwin (IIT Madras).

Hi,

Ashwin,

Can u give me more understanding about below mention steps, as AM-GM Inequality related topic still not covered in Xth grade.

Now 5t+5/t ≥ 10 --------- (AM-GM Inequality)

and t²+1/t² ≥ 2 --------- (AM-GM Inequality)

So a ≥ 12, are the values that "a" can take.

To develop a more clear understanding of the question, u need to find the minimum value which the

=5^1+x + 5^1-x + 25^x + 25^-x can have.(MINIMA OF THE EQUATION)

Calculating the minimas

For this purpose use the first let consider two separate  functions  5^1+x + 5^1-x and  25^x + 25^-x.Differentiating both the two with respect to x. =5^(1+x)log(5)-5^(1-x)log(5) and (2*5^2x*log(5))-(2*5^-2xlog5) and putting the both equal to 0.

5^(1+x)log(5)-5^(1-x)log(5)=0 ==> 5^(1+x)log(5)=5^(1-x)log(5) ==>(1+x)=(1-x) ==> 2x=0 ==> x=0

(2*5^2x*log(5))-(2*5^-2xlog5)=0 ==>(2*5^2x*log(5))=(2*5^-2xlog5) ==> 2x=-2x  ==> 4x=0 ==> x=0

As such the minimum value of 5^(1+x)log(5)-5^(1-x)log(5)= 5⁽¹⁺⁰⁾+5^(1-0)=10 and 25^x + 25^-x=25⁰ + 25^-0 =1+1=2

As such the minimum value of 5^1+x + 5^1-x + 25^x + 25^-x =10+2 =12 Ans

Hi Halak,

For the numbers x & 1/x -----------(where x is positive)

If we consider x+1/x, the minimum value it can take is 2.

And that happens when x = 1/x = 1.

The above could be obtained like this:

When the product of two numbers a,b is constant ie ab = constant, then the sum of a,b ie a+b is minimum when a = b.

Here x*1/x = 1 (a constant for all values of x)

So the sum is minimum when x = 1/x = 1.

And the value will be x + 1.x = 2.

So, in the previous solution, 5t+5/t = 5(t+1/t), the minimum is 10

And for t^2 + 1/t^2 the minimum value is 2.

And the rest follows.

Hope this helped.

Best Regards,

Ashwin (IIT Madras).