1. ## Geometric series

Hello thanks all for helping me in prep for my maths exam. I did great.

Okaydoke, but there is no rest for the wicked and i have yet hit another brick wall in Mathematics.

"List the four terms of an arithmetic progression if the last erm is ten times the first term and the sum of the terms is 121. (there are 2 possible solutions)"

2. ## Hint

Recall that arithmetic sequences are defined by having a common difference. So if the first term is a and the common difference is d, the second term will be a+d, and the third term a+2d. To find the next term we just add d, right?

So we only have four terms, making our sequence a, a+d, a+2d, a+3d. We're given two relationships between the terms. Can you turn these relationships into equations?

3. Ramujan, thanks for your hint. i did have two equations:

S(n) = 121= n/2(a + U(n))
U(n) = 10U1 = 10a

therefore

S(n) = 121 = n/2(a + 10a)
= n/2(11a)

However we have two variables. n and a. How do we get rid of these?

I know that two fine the first four terms we need to find a and d. But i can't see how to find any of them!

Please can you help me some more lol?

4. Originally Posted by mathshelpneeded
Ramujan, thanks for your hint. i did have two equations:
S(n) = 121= n/2(a + U(n))
U(n) = 10U1 = 10a
...
Hello,

(a personal remark in advance: The title of your post is really confusing...)

as Ramujan has pointed out the sum of the first four terms is:

$a+(a+d)+(a+2d)+(a+3d)=4a+6d=121$

You know that the last term is 10 times greater than the first one:

$\frac{a+3d}{a}=10 \Longleftrightarrow -9a+3d=0$

You now have to solve the system of linear equations:

$\begin{array}{r}4a+6d=121 \\-9a+3d=0\end{array}$

I don't know which method you prefer to solve this system of equation. I've got: $a=\frac{11}{2}\ \text{and } d=\frac{33}{2}$

Unfortunately I can't find a second solution.

EB

5. ## Solutions, except not sure where they came from

I noticed most of you took the first four terms to equal 121 but this is not the case. Sorry!

a = 2 or a = 1

d = (when a =2) 1.8 or d = (when a=1) 1/3

6. Originally Posted by mathshelpneeded
I noticed most of you took the first four terms to equal 121 but this is not the case. Sorry!
Originally Posted by mathshelpneeded
"List the four terms of an arithmetic progression if the last erm is ten times the first term and the sum of the terms is 121. (there are 2 possible solutions)"
If the sum of the first four terms isn't 121, then the sum of what is?

-Dan

7. The sum of all the terms. How many terms there are in the series is unknown. Hence the problem.

8. I actually figured this one out after an hour of work.

S_10 = 10/2(a_1 + a_10)

But we know that a_10 is 10 times a_1 so,

S_10 = 10/2(a_1 + 10a_1)

And we know that S_10 = 121

121 = 10/2(11a_1)

121 = 5(11a_1)

121 = 55a_1

2.2 = a_1

Which also states that

22 = a_10

Since you don't know anything else about the series and 2.2 x 1 = 2.2 = a_1 and 2.2 x 10 = 22 = a_10, then you can assume that the entire thing goes in multiples of 2.2, giving the rule:

a_n = 2.2n

So that

S_10 = 2.2 + 4.4 + 6.6 + 8.8 + 11 + 13.2 + 15.4 + 17.6 + 19.8 + 22 = 121

Thus, problem solved.

First four terms: 2.2, 4.4, 6.6, 8.8

How many terms are actually in the series is irrelevant, as long as it follows what the problem asks.

First term: 2.2
Last term: 22

2.2 x 10 = 22

10
Sigma 2.2n = 121
n=1

Thus it works.

Also, if the series starts with 2 and the common difference is 1.8, then you get the sequence:

2 + 3.8 + 5.6 + 7.4 + 9.2 + 11 + 12.8 + 14.6 + 16.4 + 18.2 + 20

2 x 10 = 20

S_11= 11/2(11a_1)

S_11 = 60.5a_1

121 = 60.5a_1

2 = a_1

Giving you the rule:

a_n = 1.8n + .2

To make sure that it fits the requirements of the problem:

First term: 2
Last term: 20

2 x 10 = 20

11
Sigma 1.8n + .2 = 121
n=1

It works as well.

First four terms: 2, 3.8, 5.6, 7.4

Thus both solutions have been found.