Question

please solve it.... question 16

Answer 1

i hope its right answer

Answer 2

Let the number of rows be 'n'.

$S_{n}=\frac{n}{2}(2a+(n-1)d)$. Here 'a' is the first term and 'd' is the common difference.

Clearly a=20 and d= -1 in this AP.

Thus,

$\frac{n}{2} (40 + (n - 1)( - 1)) = 200 \\ \\ \frac{n}{2} (40 - n + 1) = 200 \\ \\ - {n}^{2} + 41n - 400 = 0 \\ \\ {n}^{2} - 41n + 400 = 0 \\ \\ {n}^{2} - 25n - 16n + 400 = 0 \\ \\ (n - 16)(n - 25) = 0 \\ \\ n = 16 \\ n = 25$

Therefore n=16 or n=25. Finding the number of logs in 25th row gives,

$20 + (24)( - 1) \\ 20 - 24 = - 4$

This case is clearly impossible as number of logs can't be negative. Therefore the number of rows are 16.

Finding number of logs in 16th row gives,

$20 + (15)( - 1) \\ = 5$
Therefore number of rows=16 and logs in last row = 5.