200 logs are stacked in the following manner , 20 logs in the bottom row, 19 in the next row , 18 in the next to it and so on , In how may rows are the 200 logs placed and how many logs are in the top now ?

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Solution

It is an AP such that $a = 20$ & $d = - 1$
$S_{n} = 200$
$\Rightarrow 200 = \frac{n}{2} [2 (20) + (n - 1) (- 1)]$
$\Rightarrow 400 = n [40 + 1 - n]$
$\Rightarrow n^{2} - 41 n + 400 = 0$
$\Rightarrow n^{2} - 25 n - 16 n + 400 = 0$
$\Rightarrow (n - 25) (n - 16) = 0$
So $n = 16$ or $n = 25$
$T_{16} = a + 15 d = 20 - 15 = 5$
$T_{25} = a + 24 d = 20 - 24 = - 4$
Since $T_{25}$ is not possible.
$∴$ $n = 16$ & $5$ logs are placed in the top row.

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200 logs are stacked in the following manner , 20 logs in the bottom row, 19 in the next row , 18 in the next to it and so on , In how may rows are the 200 logs placed and how many logs are in the top now ?

It is an AP such that a=20 & d=−1Sn=200⇒200=n2[2(20)+(n−1)(−1)]⇒400=n[40+1−n]⇒n2−41n+400=0⇒n2−25n−16n+400=0⇒(n−25)(n−16)=0So n=16 or n=25T16=a+15d=20−15=5T25=a+24d=20−24=−4Since T25 is not possible.∴ n=16 & 5 logs are placed in the top row.