S= n/2[2a + (n - 1)d]

please solve and show the steps to get there. I'm confused at how they reached the answer the book has. Thanks

Printable View

- May 30th 2009, 12:49 PMsuddenlysmarterSolve for Variable
S= n/2[2a + (n - 1)d]

please solve and show the steps to get there. I'm confused at how they reached the answer the book has. Thanks - May 30th 2009, 02:24 PMpberardi
- May 30th 2009, 03:57 PMmr fantastic
This is the formula for the sum of the first n terms of an arithmetic series. I suggest you consult your textbook or class notes for the derivation (or Google it).

If you're attempting to apply this formula in a specific question, it would help if you actually posted that question. We are not mind readers. - May 30th 2009, 06:50 PMsuddenlysmarterSorry it was solve for a
Sorry it was solve for a and I just can't see to figure out how they achieved the answer they got. How would you approach this. In order for me to understand I would need to see how you reached your answer.

Anyone who is up for the challenge. I would greatly appreciate it. - May 30th 2009, 06:57 PMwytiaz
S = n/2[2a + (n - 1)d]

2S/n = 2a + nd - d

2S/n - nd + d = 2a

S/n - (1/2)nd + (1/2)d = a - May 30th 2009, 06:59 PMwytiaz
S= n/2[2a + (n - 1)d]

2S/n = 2a + nd - d

2S/n - nd + d = 2a

S/n - (1/2)nd + (1/2)d = a - May 30th 2009, 07:01 PMpberardi