How Do You Solve These Challenging Arithmetic Sequence Problems?

The value of ? will be different, but it will be in terms of n. So, you have an equation in n. Solve it for n.In summary, for the first question, you can use simultaneous equations to find the first term and common difference of an arithmetic progression given the sum of the first two terms and the sum of the first four terms. For the second question, you can use the formula n/2(a+l) to find the sum of an arithmetic progression, and you can remove every third term by multiplying the sum by 2/3. For the third question, you can use the same formula with a common difference of 3 to find the sum of an arithmetic progression, and then subtract the sum of the
  • #1
Timiop2008
31
0
Arithmetic Sequences - PLEASE HELP!

I would really appreciate any help to figure out the following 4 questions:

1) The Sum of the first two terms of an arithmetic progression is 18 and the sum of the first four terms is 52. Find the sum of the First eight terms

2) An arithmetic Progression has first term a and common difference -1. The sum of the first n terms is equal to the sum of the first 3n terms. Express a in terms of n.

3) Find the sum of the arithmetic progression 1,4,7,10,13,16,...,1000.
Now find the sum if every third term is removed

4) In the Sequence 1.0,1.1,1,2...,99.1,100.0 each number after the first is 0.1 greater than the previous number. Find: 1) the number of number is the sequence and 2) the sum of the numbers in the sequence.

I don't think I know any of the relevant equations (If there are any that is)
Please Help! :confused:
Thankyou
 
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  • #2


DEFINITIONS!

What is it that makes a numeric progression an arithmetic progression as opposed to, say, a geometric progession?
 
  • #3


An Arithmetic Progression is a sequence with a common difference eg 3,6,9,12,15 (with Common difference 3) whereas a Geometric Progression uses multipliers instead of a common difference i.e. x, x squared, x cubed,...
 
  • #4


OK I've done the third question using a formula I found on wikipedia: n/2(a+l) where n is the number of terms, a is the first term and l is the last term.

First I found the number of terms by dividing by the common difference:

n= 1000/d
=1000/3
=333.3
=334

Then use the formula:

n/2(a+l)
=334/2(1+1000)
=167 X 1001
=167167

and for (part 2) I just multiplied by two thirds to remove a third of the terms (every third term):
167167 x (2/3) = 111444.67
=111445

I would appreciate any help on the other Q's guys . Thanks
 
  • #5


So for problem 1, you have
a, a + d, a + 2d, a + 3d

With the given information you should be able to find a and d, and from that, the sum of the first 8 terms.
 
  • #6


Timiop2008 said:
OK I've done the third question using a formula I found on wikipedia: n/2(a+l) where n is the number of terms, a is the first term and l is the last term.

First I found the number of terms by dividing by the common difference:

n= 1000/d
=1000/3
=333.3
=334
Correct, there are 334 terms. A better way to see this is to list them (or at least list enough of them to see the pattern).
1, 1 + 1*3, 1 + 2*3, 1 + 3*3, 1 + 4*3, ... , 1 + 333*3
The ellipsis (...) means "continuing in the same pattern."
That last term is equal to 1000, and starting from the 2nd term up to the last there are 333 terms, plus the one I didn't count at the beginning, for a total of 334 terms

Timiop2008 said:
Then use the formula:

n/2(a+l)
=334/2(1+1000)
=167 X 1001
=167167
Right.
Timiop2008 said:
and for (part 2) I just multiplied by two thirds to remove a third of the terms (every third term):
167167 x (2/3) = 111444.67
=111445

I would appreciate any help on the other Q's guys . Thanks

You might be on shaky ground here. If you're off, and I don't know that you are, you won't be off by much. A better way to do it is to list the ones you're removing, say 7, 16, 25, and so on. This is also an arithmetic series, but with a different common difference. If you can list all of them (or at least a few at the beginning, an ellipsis, and the last one up around 1000, you can find its total, and subtract that from the number you got in the first part.
 
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  • #7


Thanks

I just need solutions for the first 2 Questions now.

I cannot even begin to attempt the second question (would really appreciate an help) but i tried this for the first question:

2n+d=18
4n+3d=52

So, Using simultaneous equations:

4x+3d=52
-4x+2d=36
d=16

I then put 16 into one of the equations to find x(or n) as 1.
So n=1, & d=16

For 8 terms, there will be 7 differences, so
8X1 + 7X16
=8+112
=120

Apparantly this is incorrect -the answer sheet says 168 but i don't know why!
Is there something I have overlooked?

PS. The answer to the second question is 2n-1\2 but i really don't know how to get this (I would greatly appreciate any help).
Thanks
 
  • #8


Timiop2008 said:
Thanks

I just need solutions for the first 2 Questions now.

I cannot even begin to attempt the second question (would really appreciate an help) but i tried this for the first question:

2n+d=18
4n+3d=52
The 2nd equation is wrong. You're adding four numbers, starting with n. What's the next one? The third one? The fourth one?
Timiop2008 said:
So, Using simultaneous equations:

4x+3d=52
-4x+2d=36
d=16

I then put 16 into one of the equations to find x(or n) as 1.
So n=1, & d=16

For 8 terms, there will be 7 differences, so
8X1 + 7X16
=8+112
=120

Apparantly this is incorrect -the answer sheet says 168 but i don't know why!
Is there something I have overlooked?
I get 168, also.
Timiop2008 said:
PS. The answer to the second question is 2n-1\2 but i really don't know how to get this (I would greatly appreciate any help).
Thanks

For problem 2:
2) An arithmetic Progression has first term a and common difference -1. The sum of the first n terms is equal to the sum of the first 3n terms. Express a in terms of n.
Write an expression for the first n terms.
a + (a - 1) + (a - 2) + (a - 3) + ... +(a - ?)

What should go in the place of ? so that there are n terms? Be careful when you count the terms.

Now write an expression for the first 3n terms. This expression will be equal to the one just above.
 

Related to How Do You Solve These Challenging Arithmetic Sequence Problems?

What is an arithmetic sequence?

An arithmetic sequence is a sequence of numbers where the difference between consecutive terms is constant. For example, 2, 4, 6, 8, 10 is an arithmetic sequence with a common difference of 2.

How do you find the common difference of an arithmetic sequence?

To find the common difference of an arithmetic sequence, simply subtract any term from the term that comes after it. The result will be the common difference. For example, in the sequence 3, 7, 11, 15, the common difference is 4 (7-3=4).

What is the formula for finding the nth term of an arithmetic sequence?

The formula for finding the nth term of an arithmetic sequence is:
an = a1 + (n-1)d
Where a1 is the first term, d is the common difference, and n is the term number.

What is the sum of an arithmetic sequence?

The sum of an arithmetic sequence can be found using the formula:
Sn = (n/2)(a1 + an)
Where Sn is the sum, n is the number of terms, a1 is the first term, and an is the nth term.

How can arithmetic sequences be applied in real life?

Arithmetic sequences can be applied in many real-life situations such as calculating interest rates, population growth, and financial planning. They can also be used to solve problems involving constant rates of change, such as finding the distance traveled by a moving object.

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