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Expert: - 1/27/2010


Question
1.the sum of the first 5 terms of an arithmetic series is 35 and the sum of the next 5 terms is 160. find the first term and the common difference.

2. the sum of the first 4 terms of an arithmetic series is 42 and the sum of the 3rd and 7th term is 46. find the sum of the first 29 terms.

3. the 20th term of an arithmetic series is 131 and the sum of the 6th to 10th terms inclusive is 235. find the sum of the first 20 terms.

4. the sum of 50 terms of an arithmetic series is 249 and the sum of 49 terms of the series is 233. find the 50th term of the series.  

Answer
Now to define the arithmetic series I will refer to, assume that it is
a, a+n, a+2n, a+3n, ....

1. It is given that a+n + a+2n + a+3n + a+4n a+5n = 3, so 5a + 15n = 35.
It was also given that a+6n + a+7n + a+8n + a+9n + a+10n = 160, so 5a + 40n = 160.

The 2nd equation minus the 1st is 25n = 125, so n=5.
Using the 1st or second equation, n can be put back in and the result is 5a = - 40,
so the last step is to divide both sides of that equation by 5.


2. The two equations a+n + a+2n + a+3n + a+4n = 42 and a+3n + a+7n = 46.
These work out to 4a+10n=42 and 2a+10n=46.
Subtracting te 2nd from the 1st yields 2a=-4, so a=-2.
This can be ut back in 4a+10n=42 to find n.
That can be checked to see is 2a+10n is 46.

3. The 1st equation we are given is a+20n=131.
The 2nd equation is a+6n + a+7n + a+8n + a+9n + a+10n = 235, which is 5a+40n=235.
Now the 2nd minus twice the 1st gives us 3a = -27, so a = -9.
Use the 1st equation to find n and the 2nd equation to check n and make sure it comes out right.

Given these two values, the sum of the first twenty terms is
a+n + a+2n + a+3n ... + a+19n + a+20n = 20a + 210n since 210 is the sum of 1..20.


4. Since we are given the sum of the first 50 and the sum of the first 49,
the 50th term is the difference between these two values.
That is, its 249 - 233 = 16.

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