How do I start?

J

Joyce

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Feb 16, 2006
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I know it's not too difficult but I can't seem to figure out where to begin. Any help would be appreciated.

The twelfth term of an arithmetic series is 35 and the sum of the first 20 terms is 610. Find the first five terms of the series.


Therefore:

t12=35

S20= 610



Then what should I do?
 
\(\displaystyle \L\\a_{12}=a_{1}+(n-1)d\rightarrow\\35=a_{1}+11d\)

\(\displaystyle \L\\S_{n}=\frac{n}{2}[2a_{1}+(n-1)d]\rightarrow\frac{20}{2}[2a_{1}+(20-1)d]\rightarrow\\10[2a_{1}+19d]=610\)

Solve for \(\displaystyle \L\\a_{1}\ and\ d\)

\(\displaystyle \L\\a_{1}+11d=35\)

\(\displaystyle \L\\20a_{1}+190d=610\)
 
Thank you very much for your guys help, I used the formulas and follwed through but my answer doesn't fit at the end;

a1+11d= 35

20a1+190d=610

So I inputted one formula into the other to solve for the two unknowns:

20(a1+11d-35)+190d=610

20a1+220d-700+190d=610

20a1+410d-700=610

20a1+410d=1310

At this point I just guessed and checked:

20(4)+410(3)=1310


so; d=3 and a=4

but it doesn't work the first term couldn't be four, go up by intervals of 3 and reach 35 by term 12! any help would be appreciated!
 
Joyce said:
Thank you very much for your guys help, I used the formulas and follwed through but my answer doesn't fit at the end;

a1+11d= 35

20a1+190d=610

So I inputted one formula into the other to solve for the two unknowns:

20(a1+11d-35)+190d=610

20a1+220d-700+190d=610

20a1+410d-700=610

20a1+410d=1310

At this point I just guessed and checked:

20(4)+410(3)=1310


so; d=3 and a=4

but it doesn't work the first term couldn't be four, go up by intervals of 3 and reach 35 by term 12! any help would be appreciated!
Click to expand...

Looks like you may need some practice solving systems of equations......I'm not sure AT ALL what you did with your "inputting one formula into the other."

Take the first equation and solve it for a<SUB>1</SUB>:
a<SUB>1</SUB> + 11d = 35
a<SUB>1</SUB> = 35 - 11d

Now, substitute (35 - 11d) for a<SUB>1</SUB> in the second equation:
20a<SUB>1</SUB> + 190d = 610
20(35 - 11d) + 190d = 610

You now have an equation with just one variable; solve it for d. Once you know the value of d, you can use the fact that a<SUB>1</SUB> = 35 - 11d to find a<SUB>1</SUB>

I hope this helps you.
 
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