Series and Sequences (1 Viewer)

horndog huang 2 · Jan 1, 2016

This question is from the cambridge textbook

" Logs of wood are stacked with 10 on the top row , 11 on the next , and so on . If these are 390 logs, find the number of rows and the number of logs on the bottom row "

I dont know what i did wrong :
Sn=1/2n(2a(n-1)(d)
=1/2n(2(10)+(n-1)(1))
=1/2n(20+n-1)
=1/2n(19+n)
390=19/2(n)+n^2
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InteGrand · Jan 1, 2016

horndog huang 2 said:
This question is from the cambridge textbook

" Logs of wood are stacked with 10 on the top row , 11 on the next , and so on . If these are 390 logs, find the number of rows and the number of logs on the bottom row "

I dont know what i did wrong :
Sn=1/2n(2a(n-1)(d)
=1/2n(2(10)+(n-1)(1))
=1/2n(20+n-1)
=1/2n(19+n)
390=19/2(n)+n^2
.
'.
.
.
.
'.
.
.'
'a.f
as.dasdsfvrgvaekrjfwekjfhgw ecuaew fiyw e

$Should be $\frac{n^2 }{2}$ at the end.$

horndog huang 2 · Jan 2, 2016

Still didnt get this question

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horndog huang 2 · Jan 2, 2016

Ans is 20 rows , 29 logs on bottom

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InteGrand · Jan 2, 2016

horndog huang 2 said:
Still didnt get this question

Sent from my LG-D855 using Tapatalk

$Your working was correct up to and including $S_n =\frac{n}{2} (19+n)$. We now just need to solve this quadratic equation. We have$

$\begin{align*}390 &= \frac{n}{2}(19+n)\\ \Rightarrow n(19+n) &= 780 \\ \Rightarrow n^2 +19n -780 &= 0 \\ \Rightarrow (n-20)(n + 39)&= 0.\end{align*}$

$Taking the positive solution, $n=20$, so there are 20 rows. The last row has 29 logs, since we had 10 logs in row 1 and the number of logs in each subsequent row increases by 1 (i.e. use the A.P. general term formula $t_n = t_1 + (n-1)d$ with the given values $t_1 = 10$, $d=1$, and using $n = 20$ for row 20).$

horndog huang 2 · Jan 2, 2016

Ooooooooh dam i suk, thank you for help

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Series and Sequences (1 Viewer)

horndog huang 2

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horndog huang 2

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horndog huang 2

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