Question

420 logs are stacked in a manner, one on top of the other, such that 30 logs in bottom row, followed by 29 logs in the next row, 28 logs in the row next to it, and so on. How many rows are 420 logs placed, and how many logs are there in the top most row.

Answer 1

$a=30$

$d=-1$

$s_n=420$

$420=\frac{n}{2}(2\times 30+(n-1)-1)$

$840=n(60-n+1)$

$60n-n^2+n$

$n^2=61n+840=0$

$n^2-40n-21n+840=0$

$n(n-40)-21(n-40)=0$

$n=21$ or $n=40$

$n$ con not be $40$

so, $n=21$

there are $21$ rows of logs.

$t_{21}=a+(n-1)d$

$=30+(21-1)-1$

$=30-20$

$=10$

there are $10$ logs on top row.