Linked Arithmetic progression and Harmonic progression I would like to give some introduction about the origin of my doubt and then put forth my doubt , so that people who attempt answering will know the context .
 Introduction : 
In general as speed is inversely proportional to time . Given that the distance is constant , when the velocities are in AP , the corresponding times will be in HP .Similarly when velocities are in HP , the corresponding times will be in AP . 
So for the following data , 
Person's velocity if 10 km/hr  he reaches at  1  p.m.
Person's velocity if 15 km/hr  he reaches at  11 a.m.
So for him to reach at noon what should be his speed ?
so times here are in AP 11 a.m , 12 , 1 p.m with common difference of 1 hr . so velocities will be in HP .
Speed required to reach at noon  = Harmonic mean of 10 and 15 = 12 .
So if he drives at 12 km/hr he will reach by noon . Following this method saves energy and time spent in solving this problem via equations with many unknowns .
 Problem situation : 
In a similar fashion I want to solve the following problem 
A person travels from his home to office at 4km/hr and reaches there 20 min late.If he goes at 6 km/hr he reaches 10 min early.what is the speed required to reach on time at his office?
so the 2 series under consideration are 
4km/hr    ,    6km/hr   and 
+20min    ,    -10min
Here I will not definitely be able to find it as Arithmetic mean or Harmonic mean . So my question is by moving the time series 10 minutes up  , will I be able to find the magnitude by which I should change the velocity series . If I am able to  change the AP (or HP)  and see the effect it has on the other series HP (or AP) , I can solve this in no time . Because my aim is to add 10 to time series so that the -10 will become 0 minutes (as 0 minutes delay means he is in time ) and if I know the impact it has on velocity series, I can change the velocity 6 accordingly and that will be my answer .
Note: This problem can be solved using equations of speed , time ,distance . But the motive behind using this approach is to avoid equations .Example problem avoids equations and also it is solved .
Thanks
 A: You can use your AP and HP with four terms
$$
\begin{array}{c|c|c|c|c}
\mathrm{AP} & -10 & 0 & +10 & +20 \\
\hline
\mathrm{HP} & 6=a & x=\frac{a}{1+d} & \frac{a}{1+2d} & 4 = \frac{a}{1+3d}
\end{array}
$$
then $d=1/6$ and $x=36/7$.
A: You can get your on-time speed directly from the two data points by using a (sort of) weighted HM:
$$v = \frac{20-(-10)}{\displaystyle \frac{20}{6} - \frac{-10}{4}} = \frac{36}{7}\, \text{km}/\text{hr}$$
This answer makes no assumptions about the nature of the relations between the speeds.  It merely uses the fact that the distance to work is the same in all cases.
(Truth be told, I did use a bit of algebra to come up with the expression, but the result is very straightforward.)
ADDENDUM
At the risk of alienating the OP, here is what I did to arrive at the answer above.
Let $v_1=4 \,\text{km}/\text{hr}$, $v_2=6 \,\text{km}/\text{hr}$, $\Delta_1 = 20 \, \text{min}$, $\Delta_2 = -10\, \text{min}$.  Also let $T$ be the time it should take to get to work on time.  The statement that the distance to work is the same in any case is
$$v_1 (T+\Delta_1) = v_2 (T+\Delta_2)$$
which means that
$$T = \frac{v_2 \Delta_2 - v_1 \Delta_1}{v_1-v_2}$$
Now let $v$ be the speed at which one gets to work on time.  Using the fact that the distance to work is also $v T$, we get
$$\begin{align}v T = v_1 (T+\Delta_1) \implies v &= v_1 \left ( 1 + \frac{\Delta_1}{T}\right )\\ &= v_1 \left ( 1 + \frac{\Delta_1 (v_1-v_2)}{v_2 \Delta_2 - v_1 \Delta_1}\right ) \\ &= v_1 \frac{v_2 \Delta_2 - v_1 \Delta_1 + v_1 \Delta_1 - v_2 \Delta_1}{v_2 \Delta_2 - v_1 \Delta_1} \\ &= v_1 v_2 \frac{\Delta_2-\Delta_1}{v_2 \Delta_2 - v_1 \Delta_1} \end{align}$$
or,
$$v = \frac{\Delta_1-\Delta_2}{\displaystyle \frac{\Delta_1}{v_2} - \frac{\Delta_2}{v_1}}$$
