Mean of Specific Arithmetic Means Between Two Numbers

To find the mean of the specified arithmetic means, we start by analyzing the arithmetic progression (A.P.) formed by inserting 39 arithmetic means between $a = 59$ and $b = 159$.

Let the inserted arithmetic means be $A_1, A_2, A_3, \dots, A_{39}$. The sequence $59, A_1, A_2, \dots, A_{39}, 159$ forms an A.P. with a total of $N = 39 + 2 = 41$ terms.

Let $d$ be the common difference of this A.P. The first term is $T_1 = 59$ and the $41^{\text{st}}$ term is $T_{41} = 159$. Using the formula for the $n^{\text{th}}$ term of an A.P., $T_n = T_1 + (n-1)d$: $T_{41} = 59 + (41 - 1)d = 159$ $40d = 159 - 59 = 100$ $d = \frac{100}{40} = 2.5$

The $k^{\text{th}}$ arithmetic mean $A_k$ corresponds to the $(k+1)^{\text{th}}$ term of the A.P., which is given by: $A_k = a + k \cdot d = 59 + 2.5k$

We need to calculate the mean of $A_{25}, A_{28}, A_{31},$ and $A_{36}$: $\text{Mean} = \frac{A_{25} + A_{28} + A_{31} + A_{36}}{4}$

First, let's substitute the formula for each mean: $A_{25} + A_{28} + A_{31} + A_{36} = (59 + 2.5 \times 25) + (59 + 2.5 \times 28) + (59 + 2.5 \times 31) + (59 + 2.5 \times 36)$ $= 4 \times 59 + 2.5 \times (25 + 28 + 31 + 36)$

Calculating the sum of the indices: $25 + 28 + 31 + 36 = 120$

Substituting this back into the sum: $A_{25} + A_{28} + A_{31} + A_{36} = 236 + 2.5 \times 120$ $= 236 + 300 = 536$

Now, calculating the required mean: $\text{Mean} = \frac{536}{4} = 134$

Thus, the mean of $A_{25}, A_{28}, A_{31},$ and $A_{36}$ is equal to 134.

Correct Option: D