Finding a and d from the Sum of an Arithmetic Series

[AN630078]

Homework Statement

Hello, I been revising sequences and series and have found the problem below. I normally do not struggle too much with these problems but the wording has confused me a little. I have approached question 1 in two ways but I do not think that either is necessarily correct. I would very much appreciate if anyone could comment upon my workings or offer any guidance

An arithmetic series has first term a, and common difference d. The sum of the first four terms is 139 and the sum of the next four terms is 115.

Question 1: Find the values of a and d
Question 2: Find the value of n for which the nth term is the last positive term in the series.

Relevant Equations

Sn=1/2n(2a+(n-1)d
un=a+(n-1)d

Question 1; Method 1
If the sum of the first four terms is 139 then S4=139
139=1/2(4)(2a+(4-1)d)
139=2(2a+3d)
139=4a+6d----- [1]

The part of this question that is confusing is the "the sum of the next four terms is 115".
Would this mean that S8=S4+115=139+115=254?

In which case;
254=1/2(8)(2a+(8-1)d)
254=4(2a+7d)
254=8a+28d----- [2]

Subtract [2] from 2*[1] (multiply [1] by 2 so the values of a are the same)

8a+12d=278
8a+28d=254
-16d=24
d=-1.5

When d=-1.5, substitute into [1] to find a:
4a+(6*-1.5)=139
4a-9=139
4a=148
a=37

Method 2
Again if the sum of the first four terms is 139 then S4=139
139=1/2(4)(2a+(4-1)d)
139=2(2a+3d)
139=4a+6d----- [1]

However, would the sum of the next four terms actually mean S8-S5 inclusive?
In which case 115=S8-S5
115=1/2(8)(2a+(8-1)d)-1/2(5)(2a+(5-1)d)
115=4(2a+7d)-2.5(2a+4d)

Expand the brackets:
115=8a+28d-5a-10d
115=3a+18d ----- [2]

Subtract [2] from 3*[1] (multiply by 3 so the d terms are the same)
3*139=4a+6d becomes 417=12a+18d

12a+18d=417
2a+18d=115
10a=302
a=30.2

Substitute a into [1] to find d:
4a+6d=139
(4*30.2)+6d=139
120.8+6d=139
6d=18.2
d~3.03

I think the results of my second method are more unlikely but I have seen the approach of taking a sum of terms in a series to be e.g. between the sum of the first eight terms and first five terms inclusive.

Question 2; So, I do not think I can adequately solve this until I have firmly found the values of a and d.
However, would I be looking for when the nth term of the series being un=a+(n-1)d is >0
un>0
So I would remove the inequality sign to instead solve the equation
0=a+(n-1)d
If, the common difference is indeed -1.5 and the first term 37 then:
0=37+(n-1)-1.5
-37=-1.5(n-1)
-37/-1.5=n-1
n=24.67+1
n ~ 25.67

When n= 25, u25=37+24*-1.5=1
When n=26, u26=37+25*-1.5=-0.5

So would the last positive term of the series be 1 when the nth term of the series is 25?

Do either of my solutions show potential for an appropriate approach in comprehensively solving this problem.

 

[etotheipi]

Your first approach using $S_8 - S_4$ is the correct one. It might help you to write it explicitly:$$\begin{align*}S_4 &= u_1 + u_2 + u_3 + u_4 \\ \\ S_8 &= (u_1 + u_2 + u_3 + u_4) + (u_5 + u_6 + u_7 + u_8)\\ \\ S_8 - S_4 &= u_5 + u_6 + u_7 + u_8\end{align*}$$I haven't looked through all of your algebra, but I would have thought you can just use$$ \begin{align*}S_4 = 2(2a + 3d) &= 4a + 6d = 139 \\ \\ S_8 - S_4 = 4(2a + 7d) - 2(2a + 3d) &= 4a + 22d = 115 \end{align*}$$And then subtract the two equations:$$16d = -24 \implies d = -3/2, a = 37$$

So would the last positive term of the series be 1 when the nth term of the series is 25?

Yes!

 

[PeroK]

The part of this question that is confusing is the "the sum of the next four terms is 115".

The first eight terms of an arithemetic series are:
$$a, a+d, a+2d, a+3d, a+4d, a+5d, a+6d, a+7d$$
Where is the confusion over which are the first four terms and which are the next four?