Question

200 logs are stacked in the manner that 20 logs in the bottom row, 19 in

the next row, 18 in the next to it and so on. In how many rows 200 logs are

placed and how many logs are in the top row?​

Answer 1

Answer:

In 16 rows and 5 logs are placed in 1st row

Step-by-step explanation:

Let the total logs be Sn i.e. Sn = 200

Sn : 20 + 19 + 18..........

Here, a = 20

d = 19 - 20 = -1

It is given that ,

Sn = 200

n/2[2a + (n-1)d] = 200

n/2[2*20 + (n-1)(-1)] = 200

n[40 - n + 1] = 400

n[41 - n] = 400

41n - n^ = 400

n^ - 41n + 400 = 0

n^ - 25n - 16n + 400 = 0

n(n - 25) - 16(n - 25) = 0

(n - 25) (n - 16) = 0

n - 25 = 0 OR n - 16 = 0

n = 25 OR n = 16

Let last term = l

l = a + (n - 1)d

l = 20 + (16 - 1)(-1) OR l = 20 + (25 - 1)(-1)

l = 20 + 15(-1) OR l = 20 + 24(-1)

l = 20 - 15. OR l = 20 - 24

l = 5. OR l = -4

l = -4 cannot be possible ( Since no. of logs cannot be -ive)

Therefore, l = 5

So, the number of rows in which 200 logs are stacked are 16 and the top row contains 5 logs.

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Answer 2

$\huge\sf\underline\purple{Solution :- }$

$\sf{The \: arrangement \: of \: logs} \\ \sf{20 \: , \: 19 \: , \: 18 \: ... \: forms \: an \: A.P. with \: a = 20 , d = -1} \\ \\ \sf{200 \: logs \: are \: arranged \: in \: n \: rows} \\ \\ \sf{ S_{n} = 200 } \\ \\ \sf{S_{n} = \frac{n}{2} \: [2a +(n-1) d] } \\ \\ \sf{200 = \frac{n}{2}[2 \times 20 + (n - 1)( - 1)]} \\ \\ \sf{200 \times 2 = n[40 \: - n + 1]} \\ \\ \sf{400 = n[41 - n]} \\ \\ \sf{400 = 41n - {n}^{2} } \\ \\ \sf{{n}^{2} - 41n + 400 = 0} \\ \\ \sf{ {n}^{2} - 25n - 16n + 400 = 0 } \\ \\ \sf{n(n - 25) - 16(n - 25) = 0} \\ \\ \sf{(n - 25)(n - 16) = 0} \\ \\ \sf{n - 25 = 0 \: \: \: or \: \: \: n - 16 = 0} \\ \\ \sf{n = 25 \: \: \: or \: \: \: n = 16} \\ \\ \sf{If \: \: n = 25} \\ \\ \sf{ t_{n} = a + (n - 1)d} \\ \\ \sf{ t_{25} = 20 + (25 - 1)( - 1)} \\ \\ \sf{t_{25} = 20 + 24( - 1)} \\ \\ \sf{t_{25} = 20 - 24} \\ \\ \sf{t_{25} = - 4} \\ \\ \sf{No. \: of \: logs \: in \: the \: 25th \: row \: cannot \: be \: negative}$ $\sf{n \neq 25} \\ \\ \sf{Therefore, n=16}$

$\sf{t_{n} = a + (n-1)d } \\ \\ \sf{t_{16} = 20+(16-1)(-1)} \\ \\ \sf{t_{16} = 20 + 15 (-1) } \\ \\ \sf{t_{16} = 20-15} \\ \\ \sf{t_{16} = 5}$

200 logs are placed in 16 rows and there are 5 logs in top row