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Rathviriya · Oct 4, 2009, 08:34 PM

Given a,b,c,d,e as the progression of arithmetic.Calculate a,b,c,d,e by a+b+c+d+e=125 and a+b=38.

Clough · Oct 4, 2009, 09:40 PM

Hi, Rathviriya!

If you post what you think the answer is and how you arrived at it, you'll be more likely to have someone come along who is knowledgeable to confirm whether your answer is correct, and if it isn't correct, to teach you how you might come up with the correct answer yourself.

Thanks!

Unknown008 · Oct 5, 2009, 11:54 PM

Hint: use the sum formula for arithmetic progression:

$S_n=\frac n2 (a+l)$

Sn being the sum of the first n terms.

So, that brings it to: $125 = \frac52 (a+e)$

$50 = a+e$

And $a+b = 38$

The formula for the nth term is $T_n = a+(n-1)d$

For the first term, this is 'a'.
The second term, b, it is : $a + (2-1)d= a+d$ where the 'd' here is the common difference.
The fifth term, e, is : $a + (5-1)d=a+4d$

Back the equations:

$a+b = 38$
|
V
$a+(a+d) = 38$

$50 = a+e$
|
V
$50 = a + (a+4d)$

with those two, you can solve simultaneously, and find the common difference d and the first term a. With those, the rest is easy!
Hope it helped! :)

Clough · Oct 6, 2009, 12:45 AM

Hi again, Rathviriya!

Now, you've been given the advice and help of Unknown008, who is an Expert at these sort of things.

What do you think, please?

Thanks!

mathwiz3502 · Oct 9, 2009, 04:30 AM

Or, you could find two numbers that add up to 38, but they also have to have a difference that when added to the bigger number, it would equal 25 or the mean of a, b, c, d, and e.