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### Two Arithmetic Series

```
Date: 7/1/96 at 7:27:6
From: Novica
Subject: Two Series

an and bn are two arithmetic series, An and Bn are sums of first n
elements. If (4n+27) * An = (7n+1) * Bn, compute a11/b11.

Yours truly,
Dr. Nele
```

```
Date: 7/1/96 at 11:56:16
From: Doctor Anthony
Subject: Re: Two Series

Let the 'a' series be  a, (a+d), (a+2d) etc
Let the 'b' series be  b, (b+e), (b+2e) etc

Then with n=1 in the given equation (4+27)a = (7+1)b
31a = 8b   and a/b = 8/31

An = (n/2)(2a+(n-1)d)  and Bn = (n/2)(2b+(n-1)e)  In the series of
equations connecting An and Bn we shall omit the term (n/2) as it
appears on both sides of the equation.

With n=2 we have  35(2a+d) = 15(2b+e)
With n=3 we have  39(2a+2d) = 22(2b+2e)

We now write out these equations to get:

70a - 30b + 35d - 15e = 0
78a - 44b + 78d - 44e = 0

Dividing through by b, we get for these equations

70(a/b) - 30 + 35(d/b) - 15(e/b) = 0
and      78(a/b) - 44 + 78(d/b) - 44(e/b) = 0   but we know a/b = 8/31
so
70(8/31) - 30 + 35(d/b) - 15(e/b) = 0
and      78(8/31) - 44 + 78(d/b) - 44(e/b) = 0

Collecting and simplifying terms we have:

35(d/b) - 15(e/b) = 370/31
and      78(d/b) - 44(e/b) = 740/31

These equations lead to (d/b) = 14/31
(e/b) =  8/31

Now we can evaluate a11/b11 since we know a/b = 8/31,  d/b = 14/31,
and e/b = 8/31
a11/b11 = (a+10d)/(b+10e) = (a/b + 10(d/b))/(1 + 10(e/b))

= (8/31 + 140/31)/(1 + 80/31)

= (8 + 140)/(31 + 80)

=  148/111

=  4/3

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Sequences, Series

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