Series, arithmatic progression.

  • Thread starter psyclone
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In summary, the problem states that in an Arithmetic Progression, the sum of terms p, q, and r is equal to P, Q, and R respectively. The equation to be solved is \frac{P (q - r )}{p} + \frac{Q (r - p )}{q} + \frac{R (p - q)}{r} = 0. The solution involves plugging in expressions for the sums of p, q, and r, and simplifying to arrive at the desired equation.
  • #1
psyclone
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Homework Statement


Hi All,
I found this problem,

The sum of p, q, r terms of an Arithmetic Progression, are P, Q, R respectively: show that[tex] \frac{P (q - r )}{p} + \frac{Q (r - p )}{q} + \frac{R (p - q)}{r} = 0 [/tex]

Homework Equations



3. The Attempt at a Solution [/B]

My thoughts on how to start the problem is;

if
[tex] S_{n} = \frac{a}{2} (n + (n-1)d ) [/tex]

then the sum of say 'p' terms, would be

[tex] P = S_{p} = \frac{a}{2} (p + (p-1)d ) [/tex]

Therefore;

[tex] Q = S_{q} = \frac{a}{2} (q + (q-1)d ) [/tex][tex] R = S_{r} = \frac{a}{2} (r + (r-1)d ) [/tex]

If I used the following series, to simplify a little, [tex] S_{n} = 1 + 2 + 3 ... + n, [/tex] then [tex] S_{n} = \frac{1}{2}n(n+1) [/tex]

But how to form the above equation, which combines all series, which includes all terms (i.e p q, r, P, Q & R)?
 
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  • #2
I guess you can just plug in those expressions for the sums and simplify.
 
  • #3
Solved it!

[tex] \frac{P}{p} = \frac{1}{2}(a + d(p-1)) [/tex] [tex] (eq 1) [/tex]

Use sum of Q, to let a be the subject.

[tex] a = 2\frac{Q}{q} - d(q-1) [/tex] [tex] (eq 2) [/tex]

Sub, eq 1 into eq 2

[tex] d = 2 \frac{Q}{(p-q)} ( \frac{P}{p} - \frac{Q}{q} ) [/tex]

Do the same for Sum Q & R, solve for d.

[tex] d = 2 \frac{Q}{(p-r)} ( \frac{Q}{q} - \frac{R}{r} ) [/tex]

[tex] 2 \frac{Q}{(p-q)} ( \frac{P}{p} - \frac{Q}{q} ) = 2 \frac{Q}{(p-r)} ( \frac{Q}{q} - \frac{R}{r} ) [/tex]

Simplify,

[tex] \frac{P (q - r )}{p} + \frac{Q (r - p )}{q} + \frac{R (p - q)}{r} = 0 [/tex]
 

Related to Series, arithmatic progression.

1. What is a series?

A series is a sequence of numbers or terms that are added together in a specific order.

2. What is an arithmetic progression?

An arithmetic progression is a sequence of numbers where the difference between any two consecutive terms is constant.

3. How do you calculate the sum of an arithmetic progression?

The sum of an arithmetic progression can be calculated using the formula: S = (n/2)(2a + (n-1)d), where n is the number of terms, a is the first term, and d is the common difference.

4. Can an arithmetic progression have a negative common difference?

Yes, an arithmetic progression can have a negative common difference. This means that the terms in the sequence will decrease instead of increase.

5. What is the nth term of an arithmetic progression?

The nth term of an arithmetic progression can be found using the formula: an = a1 + (n-1)d, where a1 is the first term and d is the common difference.

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