Keeps sliding!

The pole is 50m tall. Suppose that he climbs 10 times without slips must be reach 55m since 1+2+...+10=55, but if he also slips 10 times, he almost reach 45m. Until here, he takes 10x55 + 5x10 seconds = 600 seconds.
Next he have to climbs 5m more which takes 5x10 seconds = 50 seconds.
Thus the minimum time taken is 600+50 = 650 seconds

In the first step displacement is 0 In second step it is 1 m In third it is 3 m And so on..

In tenth step it is 45 m

Thus he has climbed 10 times up in the way 1+2+3+4+5+6+7+8+9+10 m Which sums up to 55.

And slipped down 10 times. He needs to travel 5m more to reach the top.

Total time taken is thus (55+5)x10 for up and 10 x 5 for down. Total is 650 seconds.

I scribbled down a cumulative time table stepwise, with achieved height on left and time-taken on right which went like 1-10, 0-15, 2-35, 1-40, 4-70, 3-75, . . . and finally reached 45-600, 50-650. Not the right way to do it. But I thought I should share it. Wisnawa has already given the right key to it.

I felt a need to write it out nicely in Latex so here goes.

We take the sum of climbing up $k-1$ after let's say 10 steps. Where $k$ represents the increasing distance the man travels up the pole before sliding down.

$\displaystyle\sum_{k=1}^{10}k-1=45$

This worked out nicely because we only have 5 more meters to go. Then we take a second sum for the amount of time this took where we solve for time rather than distance.

$\displaystyle\sum_{k=1}^{10}10k+5=600$

Because in the 11th step the man will climb 11m before he goes down 1m, he will not slide down the 1m at all since he only has 5m to go. Given $1m=10s$ we can solve for $5m=50s$ .

$\Rightarrow 600+50= 650\:seconds$

For a "general" case, consider the "steps" as: $(1-1)+(2-1)+(3-1)+(4-1) …. +(N-1)+ X = 50$ where X is the size of the final "step"

Rearranging terms yields: $(1+2+3+4+...+N) + N(-1) + X = 50$

The sum of successive N integers is given by: $\frac{N*(N+1)}{2}$ $$ so $\frac{N*(N+1)}{2} - N + X = 50$

Rearranging yields the quadratic $N^2 - N - (100-2X) = 0$

As the final "Step" must be smaller than the next logical step in the series, $X < N$

Solving the quadratic yields $(N-10)(N+9) = 0$ with $X=5$

Ignoring the negative root yields $N = 10$

Back to the original formula : $\frac{10*(10+1)}{2}\ - 10 + 5 = 50$ => $55 - 10 + 5 = 50$ => $60 - 10 = 50$

So he climbs a total of 60 metres up and slides a total of 10 metres down and the time is $60*10secs + 10*5secs = 650secs$

Please forgive me for sneaking a quadratic into a section on linear equations! :)