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250 penguins marched in rows. THere was one penguin in in the first row, 2 penguins in the second row, and 3 penguins in the third row, etch. How many rows of penguins were there? Was the last row full?

Sagot :

naǫ
You can solve it using an arithmetic sequence.

The nth term of the sequence is equal to the number of penguins in the nth row. It's equal to the number of the row.
[tex]a_n=n[/tex]

There was one penguin in the first row.
[tex]a_1=1[/tex]

The sum of the sequence:
[tex]S=\frac{n(a_1+a_n)}{2}=\frac{n(1+n)}{2}=\frac{n+n^2}{2}[/tex]

There were 250 penguins - set the sum equal to 250 and solve:
[tex]\frac{n+n^2}{2}=250 \ \ \ |\times 2 \\ n+n^2=500 \\ n^2+n-500=0 \\ \\ a=1 \\ b=1 \\ c=-500 \\ b^2-4ac=1^2-4 \times 1 \times (-500)=1+2000=2001 \\ \\ n=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-1 \pm \sqrt{2001}}{2 \times 1}=\frac{-1 \pm \sqrt{2001}}{2} \\ n=\frac{-1 -\sqrt{2001}}{2} \ \lor \ n=\frac{-1+\sqrt{2001}}{2} \\ n \approx -22.87 \ \lor \ n \approx 21.87[/tex]

The number of rows can't be a negative number so n≈21.87.
So, there were 21 full rows and one not full.
Calculate the number of penguins in 21 rows:
[tex]S_{21}=\frac{21+21^2}{2}=\frac{21+441}{2}=\frac{462}{2}=231 \\ \\ 250-231=19[/tex]
There were 19 penguins in the last row.

The answer:
There were 22 rows of penguins. The last row wasn't full, it contained 19 penguins.