# Arithmetic progression

• Jul 12th 2009, 02:43 PM
Logic
Arithmetic progression
Greetings,
I am quite curious about this problem as I have definitely failed to understand my mistake. I have used a similar strategy for an easier problem where q, r and p were known ( p = 6, q = 5, r = 0 if it helps in any way ) and it worked perfectly for that one.
• Jul 12th 2009, 10:38 PM
sa-ri-ga-ma
In arithmetic progression sum of the n terms is given by
Sn = n/2[2a + ( n - 1 )d
= na + n^2d/2 - nd/2
Compare this equation with the given equation. You get
d/2 = p or d = 2p and
a - d/2 = q or a = p + q.
In the given equation r may be zero.
• Jul 13th 2009, 09:11 AM
Logic
Yes, I know of this solution. However, what I fail to understand is why in mine I get that any r does the job whereas in yours, which is also the solution in the textbook, r is required to be 0 only.
• Jul 13th 2009, 10:54 AM
icemanfan
Quote:

Originally Posted by Logic
Yes, I know of this solution. However, what I fail to understand is why in mine I get that any r does the job whereas in yours, which is also the solution in the textbook, r is required to be 0 only.

$S_3 = S_2 + a_3$

$a_3 = S_3 - S_2 = (9p + 3q + r) - (4p + 2q + r) = 5p + q$

Since $a_2 = 3p + q$ we have that $d = 2p$.

Your logic correctly worked out that $d = 2p - r$.

However, the only way that d can be equal to both $2p - r$ and $2p$ is if $r = 0$.