MHB *Find the sum of the first 17 terms

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The discussion focuses on finding the sum of the first 17 terms of an arithmetic series starting with \(8+\sqrt{7}\) and having a common difference of \(d = -2 - \sqrt{7}\). The initial term is \(a_1 = 8+\sqrt{7}\) and the formula for the sum of the first \(n\) terms is applied. One participant calculates the sum as \(S_n = 119\sqrt{7} - 136\), while another corrects it to \(-119\sqrt{7} - 136\) after noticing a sign error. The conversation highlights the importance of careful calculation and attention to signs in arithmetic series problems. The final result is confirmed as \(-119\sqrt{7} - 136\).
karush
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Find the sum of the first $17$ terms of the arithmetic series:

$8+\sqrt{7}$, $6$, $4-\sqrt{7 }$...

$a_1=8+\sqrt{7}$; $n=17$; $d=2+\sqrt{7 }$

$\displaystyle\sum_{k=1}^{n}(a_1-kd)=136 \sqrt{7 }-170$

Don't have book answer for this?

Much Mahalo
 
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Hi karush,

$$n=17,a_1=8+\sqrt7,d=-2-\sqrt7$$

Now use

$$S_n=\dfrac{n}{2}[2a_1+(n-1)d]$$
 
$$n=17,a_1=8+\sqrt7,d=-2-\sqrt7$$
$$S_n=\frac{n}{2}[2a_1+(n-1)d]=119\sqrt{7}-136$$
 
I got $$-119\sqrt{7}-136$$.
 
your right didn't see the - sign on the TI
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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