# Arithmetic series help

• Jan 5th 2008, 12:54 PM
nugiboy
Arithmetic series help
Hi im a bit confused on this question.

-

The first 3 terms of an arithmetical progression are 7, 5.9 and 4.8.
Find
(i) the common difference
(ii) the smallest value of n for which the sum to n terms is negative

-

I worked out the common difference to be -1.1, which i know is right. Im just a bit stuck on part 2. I know the formula for the sum of terms in an arethmetic series.

Sum of n = n/2 (2a + (n-1)d)

n = term ie:5th term
a = first term, which in this case is 7
d = common difference which is -1.1

I tried to solve it by making Sum of n< 0
I get a wierd answer. Can someone show me where ive gone wrong.
• Jan 5th 2008, 02:05 PM
mr fantastic
Quote:

Originally Posted by nugiboy
Hi im a bit confused on this question.

-

The first 3 terms of an arithmetical progression are 7, 5.9 and 4.8.
Find
(i) the common difference
(ii) the smallest value of n for which the sum to n terms is negative

-

I worked out the common difference to be -1.1, which i know is right. Im just a bit stuck on part 2. I know the formula for the sum of terms in an arethmetic series.

Sum of n = n/2 (2a + (n-1)d)

n = term ie:5th term
a = first term, which in this case is 7
d = common difference which is -1.1

I tried to solve it by making Sum of n< 0
I get a wierd answer. Can someone show me where ive gone wrong.

Sum = n/2 (14 + (n - 1)(-1.1)) = n/2 (15.1 - 1.1n).

You require the smallest, non-negative, integer value of n such that sum < 0:

The graph of Sum = n/2 (15.1 - 1.1n) has an n-intercept at n = 13.7272. So round 13.7272 UP: n = 14 is the required answer. You should calculate the sum for n = 13 and then for n = 14 as a check.
• Jan 6th 2008, 12:41 AM
nugiboy
Quote:

Originally Posted by mr fantastic
Sum = n/2 (14 + (n - 1)(-1.1)) = n/2 (15.1 - 1.1n).

You require the smallest, non-negative, integer value of n such that sum < 0:

The graph of Sum = n/2 (15.1 - 1.1n) has an n-intercept at n = 13.7272. So round 13.7272 UP: n = 14 is the required answer. You should calculate the sum for n = 13 and then for n = 14 as a check.

Thanks for the reply. Im a bit confused on how you got the n - intercept as 13.7272 from Sum = n/2 (15.1 - 1.1n). Could you please explain.
• Jan 6th 2008, 12:48 AM
mr fantastic
Quote:

Originally Posted by nugiboy
Thanks for the reply. Im a bit confused on how you got the n - intercept as 13.7272 from Sum = n/2 (15.1 - 1.1n). Could you please explain.

Thinking instead of the graph of y = x/2 (15.1 - 1.1 x) (where I've relabelled sum as y and n as x) might help. To get the x-intercept, sub y = 0:

0 = x/2 (15.1 - 1.1 x).

From the null factor law (it goes by other names but that's the name I usually use) either:

x/2 = 0 => x = 0, or

15.1 - 1.1 x = 0 => 15.1 = 1.1 x => x = 15.1/1.1 = 13.7272 (to 4 dp).

Capisce?
• Jan 6th 2008, 01:07 AM
nugiboy
Quote:

Originally Posted by mr fantastic
Thinking instead of the graph of y = x/2 (15.1 - 1.1 x) (where I've relabelled sum as y and n as x) might help. To get the x-intercept, sub y = 0:

0 = x/2 (15.1 - 1.1 x).

From the null factor law (it goes by other names but that's the name I usually use) either:

x/2 = 0 => x = 0, or

15.1 - 1.1 x = 0 => 15.1 = 1.1 x => x = 15.1/1.1 = 13.7272 (to 4 dp).

Capisce?

Thanks man! I get it now. I think i just got mixed up with the factorising.(Whew)
• Jan 6th 2008, 07:01 AM
chinkmeista
(Bear)Ehew...govind Jup.