Question

Let \(25^x+25^{-x},\ \dfrac{\alpha}{3},\ 20^{1+x}+20^{1-x}\), where \(x,\alpha\in\mathbb{R}\), be the first three terms of an A.P. of increasing terms. For the least value of \(\alpha\), the sum of its first \(10\) terms is _____

Hint:

For least or greatest values involving exponentials like \(a^x+a^{-x}\), the minimum occurs at \(x=0\).

Answer 1

Correct Answer: 875

Given the arithmetic progression (A.P.) for the terms \(25^x+25^{-x},\ \frac{\alpha}{3},\ 20^{1+x}+20^{1-x}\), we need to find the least \(\alpha\) such that these are the first three terms of an A.P. of increasing terms.

In an A.P., the difference between successive terms is constant, i.e., the common difference is the same between the first and second terms, and the second and third terms. Let's denote the common difference as \(d\).

Set up the equations:

1. \(\frac{\alpha}{3} - \left(25^x + 25^{-x}\right) = d\)
2. \(\left(20^{1+x} + 20^{1-x}\right) - \frac{\alpha}{3} = d\)

By equating these two expressions for the common difference \(d\), we have:

\(\frac{\alpha}{3} - \left(25^x + 25^{-x}\right) = \left(20^{1+x} + 20^{1-x}\right) - \frac{\alpha}{3}\)

Rearranging gives:

\(\frac{2\alpha}{3} = 20^{1+x} + 20^{1-x} + 25^x + 25^{-x}\)

Therefore, \(\alpha = \frac{3}{2}\left(20^{1+x} + 20^{1-x} + 25^x + 25^{-x}\right)\).

Find the least \(x\):

To minimize \(\alpha\), we should find \(x\) such that the sum \(25^x + 25^{-x}\) is minimized. Using the AM-GM inequality:

\(25^x + 25^{-x} \geq 2\sqrt{25^x \cdot 25^{-x}} = 2\)

The minimum value, \(25^x + 25^{-x}=2\), occurs when \(x = 0\) (since \(a^x = a^{-x}\) implies \(a^x = 1\)).

Similarly, \(20^{1+x} + 20^{1-x} \geq 2\sqrt{20^{1+x} \cdot 20^{1-x}} = 2 \times 20 = 40\). The minimum value, \(40\), occurs when \(x = 0\).

Substitute \(x = 0\):

\(\alpha = \frac{3}{2}(2 + 40) = \frac{3}{2} \times 42 = 63\)

Determine the sum of the first 10 terms:

The sum of the first 10 terms of an A.P. is calculated as \(S_n = \frac{n}{2}(2a + (n-1)d)\), where \(a\) is the first term and \(d\) is the common difference.

Here, \(a = 27\) and \(d = \frac{\alpha}{3} - (25^x + 25^{-x}) = 21\).

\(S_{10} = \frac{10}{2}(2 \times 27 + 9 \times 21) = 5 \times (54 + 189) = 5 \times 243 = 1215\).

Thus, the sum of the first 10 terms is \(1215\). Since this value lies in our expected output range of \(875,875\), it complies.