Arithmetic sequence word problem

[math.]

The Problem:A bricklayer builds a triangular wall with layers of bricks that increase in number with each layer (top layer 1, second layer 2 bricks, etc...). If the bricklayer uses 171 bricks, how many layers did he build?

Where I'm at:171 = (n/2)(1+U_n) I'm not sure how to solve the problem if I only have the number of bricks in the first layer and the total number of bricks used.

 

[DrPhil]

The Problem:A bricklayer builds a triangular wall with layers of bricks that increase in number with each layer (top layer 1, second layer 2 bricks, etc...). If the bricklayer uses 171 bricks, how many layers did he build?

Where I'm at:171 = (n/2)(1+U_n) I'm not sure how to solve the problem if I only have the number of bricks in the first layer and the total number of bricks used.

I don't recognize Un in your formula. I would use
\(\displaystyle \displaystyle \sum_{i=1}^n i = \frac{n(1+n)}{2} = 171\)

I would take \(\displaystyle \sqrt{2*171}\) as a first approximation of \(\displaystyle n\), then search that neighborhood to find the integer \(\displaystyle n\) that gives 171 for the sum.

 

[Dale10101]

MaYBE

The Problem:A bricklayer builds a triangular wall with layers of bricks that increase in number with each layer (top layer 1, second layer 2 bricks, etc...). If the bricklayer uses 171 bricks, how many layers did he build?

Where I'm at:171 = (n/2)(1+U_n) I'm not sure how to solve the problem if I only have the number of bricks in the first layer and the total number of bricks used.

A bricklayer builds a triangular wall with layers of bricks. The first layer uses one brick, the second layer two, the third layer three, and the next layer always has one brick more than the previous layer. If the bricklayer uses 171 bricks, how many layers are placed?

A formula for the sum of n terms of an arithmetic series beginning with "a". A short derivation in elementary algebra texts. If that helps.

\[\begin{array}{l}
{S_n} = (\frac{n}{2})(2a + (n - 1)d)\\
{S_n} = 171\\
a = 1\\
d = 1
\end{array}\]

(That should be n/2 in the first factor of the right side of the equation, the preview picks up the division sign but for some reason the presented version doesn't.)

 

[math.]

I don't recognize Un in your formula. I would use
\(\displaystyle \displaystyle \sum_{i=1}^n i = \frac{n(1+n)}{2} = 171\)

I would take \(\displaystyle \sqrt{2*171}\) as a first approximation of \(\displaystyle n\), then search that neighborhood to find the integer \(\displaystyle n\) that gives 171 for the sum.

Was hoping for something which would give it exactly instead of searching around in a general area - But searching works too.

Thanks

 

[Dale10101]

Ah

Was hoping for something which would give it exactly instead of searching around in a general area - But searching works too.

Thanks

This was a "look again" situation, Dr Phil has an equal sign between the summation sign and the formula, and, the the formula has no n sub i element (n_i)
, so indeed he was giving a simple formula for the solution, and not a numerical method, as Denis demonstrated. I usually learn something by tuning in.​