Really need to know!

Joyce

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Feb 16, 2006
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I just wrote a math test and could not get that last question. It would be greatly appreciated if someone could show me what I did wrong just so that I know for next time.


Question: In an arithmetic sequence you have twenty terms. The sum of the 20 terms is 1590. Term 8 is 62. How many of the twenty terms are below 400?


I wrote;

a= ?

d=?

t8= 62 sn= n(a+tn)/2

1590= 62(a+8)/2

a= 43.29...
(which I rounded to 43). I know know what the swequence starts with.

I then inputed everything into the arithmetic formula tn= a+(n1)d and solved for d (which i got 2.7) then I concluded that term 20 was about 94 so all twenty of the terms were under 400. I know it's completely wrong though!
 
The terms of the arithmetic squence are \(\displaystyle \L
a_n = a_1 + (n - 1)d\).
Thus
\(\displaystyle \L
\begin{array}{l}
a_8 = a_1 + (7)d = 62\quad \& \\
\sum\nolimits_{k = 1}^{20} {a_1 + (k - 1)d} = 20a_1 + \frac{{(19)(20)}}{2}d = 20a_1 + 190d = 1590 \\
\end{array}\).

Now solve for a<SUB>1</SUB> and d.
 
Can you solve this system of equations?
\(\displaystyle \L
\begin{array}{l}
a_1 + 7d &=& 62 \\
20a_1 + 190d &=& 1540 \\
\end{array}\)

In that system, a<SUB>1</SUB> is the first term of the arithmetic sequence and d is the common difference. Do you know that any arithmetic sequence is uniquely determined by its first term and its common difference?

Given any arithmetic sequence with a<SUB>1</SUB> as its first term and common difference d the here is the way we find any term a<SUB>n</SUB>.
\(\displaystyle \L
a_n = a_1 + (n - 1)d\).
So the eight term is a<SUB>8</SUB>= a<SUB>1</SUB>+(8−1)d.

The sum of the first N terms is:
\(\displaystyle \L
S_N = \left( {Na_1 + \frac{{\left( {N - 1} \right)N}}{2}d} \right)\).
So the sum of the first 20 terms is
\(\displaystyle \L
\left( {20a_1 + \frac{{20(19)}}{2}d} \right)\).
 
o.k that makes sense now but how would I isolate to solve for a and d?

:?
 
Perhaps someone else will do this problem for you.
I will not! After all, I completely set the problem up for you.
You, yourself, should learn how to solve a system of equations.
 
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