Number sequences

bumblebee123

Junior Member
Joined
Jan 3, 2018
Messages
200
After many attempts, I still can't get the right answer to this question. Can anyone tell me where I'm going wrong?

Question: the 12th term of an arithmetic series is 27
the 18th term of the arithmetic series is 45

a) find the sum of the first 60 terms of this arithmetic series
b) find the three consecutive terms in this arithmetic series whose sum is the same as the sum of the first 60 terms of the series

I started off by using simultaneous equations to find the value of the common difference ( d = 3 ) and the first term ( a = -6 ).

I plugged this into the formula for the sum of the first 60 terms in the series: Sn = n/2 [ 2a + ( n - 1 )d ]
S60= 60/2 [ 12 + 177 ]
S60= 30 x 189 = 5670
however this does not give me the right answer. The correct answer is: 4950

I haven't even started on the second question! does anyone know what to do? any help would be really appreciated :)
 
I started off by using simultaneous equations to find the value of the common difference ( d = 3 ) and the first term ( a = -6 ).

I plugged this into the formula for the sum of the first 60 terms in the series: Sn = n/2 [ 2a + ( n - 1 )d ]
S60= 60/2 [ 12 + 177 ]
S60= 30 x 189 = 5670
however this does not give me the right answer. The correct answer is: 4950

You used a=6 instead of a=-6 as previously stated! Fix that, and you have the right answer. Good work ...
 
Now that I have the answer to a) I have no idea how to start part b). I've never had a question like this before, can anyone give me some clues?
 
If \(a\) is the first term in the sequence, and \(d\) the common difference, then the \(n\)th term is:

[MATH]a_n=a+(n-1)d[/MATH]
Having found that \(a=-6,\,d=3\) and \(S_{60}=4950\)

I would then state:

[MATH]a_{n}+a_{n+1}+a_{n+2}=S_{60}[/MATH]
[MATH](-6+3(n-1))+(-6+3(n))+(-6+3(n+1))=4950[/MATH]
Solve this for \(n\), and then you'll know the 3 requested terms of the sequence.
 
I've used the formula for the nth term = a + ( n-1 )d
nth term = -6 + 3n - 3
nth term = -9 + 3n
4950 = -9 + 3n
4959 = 3n
n = ( 4959/ 3 ) = 1647

so the first of the three consecutive terms is 1647

If I add the common difference ( 3 ) I can find the next two terms ( 1650 and 1653 ). This is the right answer, but have I worked it out right?
 
Last edited:
Solving the equation I posted, I get:

[MATH]n=552[/MATH]
And so the 552nd term of the sequence is:

[MATH]a_{552}=-6+(551)3=1647\quad\checkmark[/MATH]
However, I don't see how you got there...\(\frac{4595}{3}\ne1647\) and \(\frac{4959}{3}\ne1647\). I don't see where you added 3 consecutive terms either. It appears that you were trying to find which single terms of the sequence is equal to the sum of the first 60.
 
Solving the equation I posted, I get:

[MATH]n=552[/MATH]
And so the 552nd term of the sequence is:

[MATH]a_{552}=-6+(551)3=1647\quad\checkmark[/MATH]
However, I don't see how you got there...\(\frac{4595}{3}\ne1647\) and \(\frac{4959}{3}\ne1647\). I don't see where you added 3 consecutive terms either. It appears that you were trying to find which single terms of the sequence is equal to the sum of the first 60.

sorry that was a typo! I should have written 4959 / 3 = 1647
 
sorry that was a typo! I should have written 4959 / 3 = 1647
Even so, as MarkFL pointed out, you find the single term (lucky it existed) that equaled S60. But you were ask to find 3 terms (3 consective terms) whose sum equaled S60. Do you see that?? If not, then let us know.
 
Question: the 12th term of an arithmetic series is 27
the 18th term of the arithmetic series is 45
a) find the sum of the first 60 terms of this arithmetic series
b) find the three consecutive terms in this arithmetic series whose sum is the same as the sum of the first 60 terms of the series
Here are lecture notes on arithmetic series. Let \(\displaystyle A\) equal the first them of the sequence.
Let \(\displaystyle d\) be the common difference in the terms of the sequence.
Then each term of the sequence is \(\displaystyle A_n=A+(n-1)d~.\)
From that simple setup, is it clear that \(\displaystyle A_1=A~?\)
Moreover, the sum of the first \(\displaystyle n\) terms \(\displaystyle S_n\) is \(\displaystyle S_n=nA+d\left(\dfrac{(n-1)n}{2}\right)\)

Now turning to your particular question,
\(\displaystyle \begin{align*}A_{12}&=A+(12-1)d=27\\A_{18}&=A+(18-1)d=45\\6d&=18\\d&=3~\&~A=-6 \end{align*}\)

Now for the Sum: \(\displaystyle \begin{align*}S_{30}&=30(-6)+3\left(\dfrac{(29)(30)}{2}\right) \\ \end{align*}\)



 
here's what I tried to do:

a = first term ( -6 )
d = common difference ( 3 )
n= term number ( ? )
nth term = sum of first 60 terms ( 4950 )

I used the formula: nth term = a + ( n - 1 )d

4950 = -6 + 3n - 3
4950 = -9 + 3n
4959 = 3n
n = 1653 ( I think I wrote this wrong before as well )

this doesn't make much sense anymore- I'm not sure what I was trying to do.

If I try using another method ( like MarkFL suggested ) :

( a + ( n - 1 ) ) + ( a + (n) ) + ( a + ( n + 1 )) = 4590

(−6+3(n−1))+(−6+3(n))+(−6+3(n+1)) = 4950

( -9 + 3n ) + ( -6 + 3n ) + ( -3 + 3n ) = 4950

- 18 + 9n = 4950
9n = 4968
n = 4968 / 9 = 552

now that n = 552 I can find the terms using the formula: nth term = a + ( n - 1 )d

a 552 = -6 + (551)3 = -6 + 1653 = 1647

If I keep adding the common difference ( 3 ) I get the next two terms: 1650, 1653

so the answer is: 1647, 1650, 1653 which is correct

Is this the method I should have used?
 
It's a method you should have used. There are others.

Here's what I did: You want three consecutive terms whose sum is 4950. The sum of 3 consecutive terms is 3 times their average, which is the middle term. So the middle term has to be 4950/3 = 1650; the others are 1647 and 1653, by subtracting and adding 3.

The problem was not so much that you used the wrong method, as that you solved the wrong problem! You made the one term, rather than the sum of three terms, equal to 4950. And you found n, not the term itself, for that wrong problem.
 
Top