Let $$A_1, A_2, \ldots, A_{39}$$ be 39 arithmetic means between the numbers 59 and 159. The mean of $$A_{25}, A_{28}, A_{31} and  A_{36}$$ is equal to :

Let the required arithmetic progression be
  $$59,\,A_1,\,A_2,\ldots,\,A_{39},\,159$$.

Total number of terms = $$39$$ means $$+\,2$$ end terms $$= 41$$.
Therefore, the common difference $$d$$ is obtained from
  $$a_{41}=a_1+40d$$, so
  $$159=59+40d \;\Longrightarrow\; d=\frac{159-59}{40}=2.5$$.

General term of the progression:
  $$A_k = 59 + k\,d = 59 + 2.5k$$, where $$k=1,2,\ldots,39$$.

Compute the required four means:

$$A_{25}=59+2.5(25)=59+62.5=121.5$$
$$A_{28}=59+2.5(28)=59+70=129$$
$$A_{31}=59+2.5(31)=59+77.5=136.5$$
$$A_{36}=59+2.5(36)=59+90=149$$

Sum of these four terms:
  $$121.5+129+136.5+149=536$$.

Their arithmetic mean is therefore
  $$\frac{536}{4}=134$$.

Option D which is: $$134$$