The problem involves an arithmetic progression (A.P.) where the sixth term \(a_6 = 2\). We need to find the common difference \(d\) that maximizes the product of three terms: \(a_1 \times a_4 \times a_5\).
Let's analyze the problem step-by-step:
After testing the values, choosing \(d = \frac{8}{5}\), you find that it results in equalizing terms and maximizing the product due to symmetry. Therefore, the common difference \(d\) is \(\frac{8}{5}\).
The sixth term of an A.P. can be expressed as:
\(a_6 = a + 5d = 2\)
were \(a\) is the first term and \(d\) is the common difference. Therefore, we have:
a = 2 - 5d
The product \(a_1 a_4 a_5\) can be expressed as:
\(a_1 a_4 a_5 = a(a + 3d)(a + 4d)\)
Substituting \(a = 2 - 5d\) into this expression, we get:
\(a_1 a_4 a_5 = (2 - 5d)(2 - 2d)(2 - d)\)
To find the maximum value of this product, we can analyze the behavior of the function:
\(f(d) = (2 - 5d)(2 - 2d)(2 - d)\)
After taking the derivative and setting it to zero, the solution in the image calculates critical points and finds that \(d = \frac{8}{5}\) maximizes the product.
So, the correct option is: \(d = \frac{8}{5}\).
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,