Arithmetic progression problem

[Averagepunter]

The sum of the first three terms of an arithmetic progression is 33. The sum of the next four terms is 128.
What is the sum of
(a) the next five terms?
(b) all the terms less than 1000?

 

[Deleted member 4993]

The sum of the first three terms of an arithmetic progression is 33. The sum of the next four terms is 128.
What is the sum of
(a) the next five terms?
(b) all the terms less than 1000?

Please share your work/thoughts about this assignment.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING

Hint: The sum of first n terms of an arithmetic series whose first term is "a" and each term differs from the next term by an amount "d"

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

 

[Averagepunter]

Please share your work/thoughts about this assignment.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING

Hint: The sum of first n terms of an arithmetic series whose first term is "a" and each term differs from the next term by an amount "d"

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

Hi Subhotosh,
Thank you for that hint. I worked on that. After extensive work, I found out that my a (first term in the series) is -1 (negative 1) and difference is: 12.
so first three terms are -1, 11, 23 and 4 terms after that are: 35, 47, 58, 70. (that add up to: 210. however the question says: that should be 128?
I am confused... Please help.
Thanks.

 

[MarkFL]

If \(a\) is the first term, and \(d\) is the difference, then from the problem statement, we may write:

[MATH]a+(a+d)+(a+2d)=33\implies a+d=11[/MATH]
[MATH](a+3d)+(a+4d)+(a+5d)+(a+6d)=128\implies 2a+9d=64[/MATH]
What do you get upon solving this system?

 

[Averagepunter]

If \(a\) is the first term, and \(d\) is the difference, then from the problem statement, we may write:

[MATH]a+(a+d)+(a+2d)=33\implies a+d=11[/MATH]
[MATH](a+3d)+(a+4d)+(a+5d)+(a+6d)=128\implies 2a+9d=64[/MATH]
What do you get upon solving this system?

I found that 'a' (first term in series) should be 5 and d (difference) should be: 6.

That works, since: 5 + 11 + 17 = 33

and next 4 terms, which are: 23 + 29 + 35 + 41 = 128

So, the answer to "(a) the next five terms" is: 47 + 53 + 59 + 65 + 71 + 77 = 372.

However, I am struggling to form an equation for the question "(b) sum of all the terms less than 1000".

Would you kindly help. Thank you in advance.

 

[pka]

I found that 'a' (first term in series) should be 5 and d (difference) should be: 6.

Correct. each term is \(\displaystyle a_n=5+6(n-1)\) So what is the value of \(\displaystyle n\) if \(\displaystyle a_n<1000~?\)

 

[Steven G]

Hi Subhotosh,
Thank you for that hint. I worked on that. After extensive work, I found out that my a (first term in the series) is -1 (negative 1) and difference is: 12.
so first three terms are -1, 11, 23 and 4 terms after that are: 35, 47, 58, 70. (that add up to: 210. however the question says: that should be 128?
I am confused... Please help.
Thanks.

Hi Subhotosh,
Thank you for that hint. I worked on that. After extensive work, I found out that my a (first term in the series) is -1 (negative 1) and difference is: 12.
so first three terms are -1, 11, 23 and 4 terms after that are: 35, 47, 58, 70. (that add up to: 210. however the question says: that should be 128?
I am confused... Please help.
Thanks.

For the record, your numbers 35, 47, 58, 70 do NOT all differ by 12. 47+12 = 59 NOT 58. Adding 12 to an integer does not change the parity of the number. Since the 1st number 35 is odd, then all the following numbers will be odd.

 

[Averagepunter]

For the record, your numbers 35, 47, 58, 70 do NOT all differ by 12. 47+12 = 59 NOT 58. Adding 12 to an integer does not change the parity of the number. Since the 1st number 35 is odd, then all the following numbers will be odd.

That was very silly of me to make that error. I appreciate you pointing it out. I will be careful next time. Thank you.

 

[pka]

For the record, your numbers 35, 47, 58, 70 do NOT all differ by 12. 47+12 = 59 NOT 58. Adding 12 to an integer does not change the parity of the number. Since the 1st number 35 is odd, then all the following numbers will be odd.

That was very silly of me to make that error. I appreciate you pointing it out. I will be careful next time.

Correct. each term is \(\displaystyle a_n=5+6(n-1)\) So what is the value of \(\displaystyle n\) if \(\displaystyle a_n<1000~?\)

I do not understand the first to quotes above, The first term is \(\displaystyle 5\) and the common difference is \(\displaystyle 6\).
The general terms are \(\displaystyle a_n=5+6(n-1)\) so that if \(\displaystyle n\le 165\) then \(\displaystyle a_n\le 1000\).
HERE is the summation.

 

[Steven G]

I do not understand the first to quotes above, The first term is \(\displaystyle 5\) and the common difference is \(\displaystyle 6\).
The general terms are \(\displaystyle a_n=5+6(n-1)\) so that if \(\displaystyle n\le 165\) then \(\displaystyle a_n\le 1000\).
HERE is the summation.

In post#3 the OP incorrectly claimed that d=12 and the 4th-7th terms were 35, 47, 58, 70. I just pointed out a quick way to see that these numbers did not all differ by 12---as their parity changed

 

[MarkFL]

I found that 'a' (first term in series) should be 5 and d (difference) should be: 6.

That works, since: 5 + 11 + 17 = 33

and next 4 terms, which are: 23 + 29 + 35 + 41 = 128

So, the answer to "(a) the next five terms" is: 47 + 53 + 59 + 65 + 71 + 77 = 372.

However, I am struggling to form an equation for the question "(b) sum of all the terms less than 1000".

Would you kindly help. Thank you in advance.

As has been posted by others, I would then continue to find the \(n\)th term. We know that:

[MATH]a_1=5[/MATH]
And:

[MATH]a_{n+1}-a_n=6[/MATH]
Without delving into linear difference equations, we can "see" that the solution is of the form:

[MATH]a_n=6n+c_1[/MATH]
We can use the first value (initial conditions) to determine the value of the parameter \(c_1\):

[MATH]a_1=6+c_1=5\implies c_1=-1[/MATH]
Hence:

[MATH]a_n=6n-1[/MATH]
So, let's find the partial sum:

[MATH]S_n=\sum_{k=1}^n(6k-1)=6\cdot\frac{n(n+1)}{2}-n=3n(n+1)-n=n(3n+2)[/MATH]
Now, we want to determine the greatest element which is less than 1000:

[MATH]6n-1\le1000[/MATH]
[MATH]n\le\frac{1001}{6}\implies n=166[/MATH]
And so we need to find the sum:

[MATH]S_{166}=166(3(166)+2)=83000[/MATH]