# An accelerating rocket rises 10m in the first second, 40m in the second second, and 70m in the third second. If the arithmetic sequence continues, how high will the rocket be after 20 seconds?

The height of the rocket after 20 seconds will be 5,900m.

We are given that the rocket rises 10m in the first second, an additional 40m in the second second, and additional 70m in the third second, and that the increases continue to follow the arithmetic sequence. We are asked to determine the height of the rocket after 20 seconds.

The increases of height are in arithmetic progression with first term `a_1=10` and a common difference of 30m. The formula for the `n^(th)` term of an arithmetic sequence is `a_n=a_1+(n-1)d` , so the `20^(th)` term is:

`a_(20)=a_1+(20-1)(30)=580`

The sum of the first n terms of an arithmetic sequence is given by

`S_n=n(a_1+a_n)/2`

The height of the rocket at t=1sec is 10m
The height of the rocket at t=2sec is 10+40=50m
The height of the rocket at t=3sec is 10+40+70=120m

The height of the rocket at t=20 seconds is `S_(20)=20(10+580)/2=5,900"m"`

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