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

Concept:
For three consecutive terms of an A.P., the middle term is the arithmetic mean of the other two.
For increasing A.P., the common difference must be positive.
The sum of the first \(n\) terms of an A.P. is \[ S_n=\frac{n}{2}\,[2a+(n-1)d]. \]
Step 1: Use the A.P. condition Let \[ T_1=25^x+25^{-x},\quad T_2=\frac{\alpha}{3},\quad T_3=20^{1+x}+20^{1-x}. \] For an A.P., \[ 2T_2=T_1+T_3 \] \[ \Rightarrow \alpha=\frac{3}{2}\Big(25^x+25^{-x}+20^{1+x}+20^{1-x}\Big). \]
Step 2: Find the least value of \(\alpha\) The expression \[ 25^x+25^{-x}+20^{1+x}+20^{1-x} \] is minimum at \(x=0\). Thus, \[ \alpha=\frac{3}{2}\Big(2+40\Big)=\frac{3}{2}\cdot42=63. \]
Step 3: Determine the A.P. \[ T_1=2,\quad T_2=\frac{63}{3}=21 \] \[ d=21-2=19. \]
Step 4: Find the sum of the first 10 terms \[ S_{10}=\frac{10}{2}\,[2(2)+9(19)] =5(4+171)=5\times175=875. \]
Final Answer: \(\boxed{875}\)