Clock losing time puzzle The question goes as:

A wall clock and a Table clock are set to correct time today on 10 pm. The wall clock loses 3 minute in 1st hour, 6 minutes in the second hour and 9 minutes in the third hour and so on. The table clock loses 5 minutes in the 1st hour, 10 minutes in the second hour and 15 minutes in the third hour and so on. When will they show the same time?

My approach:
In the first hour, the difference between the two clocks would be $2$ (obtained from $5-3$) minutes.
In the second hour, it'll be four minutes and so on. This would form an arithmetic progression with $a$ = 2 and $d = 2$. I, then, formulated the problem as:
$$2 + 4 + 6+ 8 + \dots + n = 720 $$
The RHS is $720$ because I assumed they'll meet after 12 hours.
With this, I got the root as $23.337$ hours, so I arrived at the answer as $10 \, \text{PM} + 23.337$ hours i.e $9:20 \, \text{PM} $.
Is this correct?  
EDIT: I realised this equation won't give an integral answer, and we need one as $n$ on the LHS represents the number of terms. So instead of that, I wrote it as:
$$2 + 4 + 6 + \dots + n = 720 \times k$$ where $k \in (1,2,3,4, \dots)$.
Using this method, for $k = 9$, I get the value of $n$ $\text{as}$ $80 \, \text{hours}$.
Does this seem correct?
 A: 
With this, I got the root as 23.337 hours, so I arrived at the answer as 10PM+23.337 hours i.e 12:20AM.

23.33333.. is 2/3 of an hour less than a day, or 40 min less than a day. So 10PM+23.3333... is 10PM+1day-40 min, or 9:20PM.
There are several different interpretations of this problem. For one thing, "lose a minute" is ambiguous. It could mean "be 1 minute behind" or "be 1 more minute behind". Under the first interpretation, the difference increases by 2 minutes each hour, so it will take 30 hours to differ by an hour, and 360 hours, or 15 days, to differ by 12 hours. There is now a further ambiguity as to what type of clocks they are; if they are 24-hour clocks, then it will take 720 hours, or 30 days, to show the same time.
If "lose a minute" means "be 1 more minute behind", then we need to make further assumptions. We can fit a quadratic equation to the data points given: after one hour, the wall clock shows 57 minutes. After 2 hours, it shows 111 minutes, etc. That gives the points (1,57), (2,11), (3,168). If x is the actual time in hours, and y is the shown time in minutes, then $y = 60x-\frac{3x(x+1)}2$ fits these data points, but if the problem expects us to therefore conclude that this is the right equation, it is asking us to make something of a leap. Similarly, the equation $y = 60x-\frac{5x(x+1)}2$ fits the table clock. The difference is then $\frac{2x(x+1)}2$, or just $x(x+1)$. Solving for $x(x+1) = 720$, we get:
$x(x+1) = 720$
$x^2+x-720=0$
This gives the solutions 26.3375 and −27.3375. The negative solution corresponds to some past time when they showed he same time, so we should take the positive one. This corresponds to 24+2.3375 hours, or 1 day, 2 hours, and .3375 hours. .3375 hours is 20.25 minutes. .25 minutes is 15 seconds. So this corresponds to 1 day, 2 hours, 20 minutes, and 15 seconds. Adding this to the initial time of 10PM gets to 12:20:15 AM on the second day (the day after the day after the clocks are set).
This matches the time you got, so perhaps your number 23.337 was a typo.
