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Question 1171677: Question: The sum of 40 terms of a certain arithmetic sequence is 430, while the sum of 60 terms is 945. Determine the nth term of the arithmetic sequence.
What I did so far: a1 + 39d = 430 ……….Eq(1)
a1 + 59d = 945 ……… Eq (2)
Solving for d: 20d = 515 and d = 25.75
Therefore the nth term of the arithmetic sequence= a1 +(n - 1)25.75
The different answer given in the book is: (n + 1)/2
What am I missing?
Please help.
Thanks
PLEASE FORGET THIS QUESTION: I USED THE LAST TERM FORMULA AND EQUATED THAT TO SUM!
Found 2 solutions by math_tutor2020, muthbab@gmail.com:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
I think you meant to say "the sum of the first 40 terms is 430" and "the sum of the first 60 terms is 945". Without those "first"s in there, the instructions are too vague.
If that assumption is true, then we use this formula to calculate the sum of the first n terms of an arithmetic sequence
Sn = (n/2)*(a1+an)
where,
Sn = the sum of the first n terms
n = number of terms
a1 = first term
an = nth term
If the sum of the first n = 40 terms is 430, then
Sn = (n/2)*(a1+an)
S40 = (40/2)*(a1+a40)
S40 = 20*(a1+a40)
430 = 20*(a1+a40)
430/20 = a1+a40
21.5 = a1+a40
a1+a40 = 21.5
Also, if the sum of the first n = 60 terms is 945, then,
Sn = (n/2)*(a1+an)
S60 = (60/2)*(a1+a60)
945 = (60/2)*(a1+a60)
945 = 30*(a1+a60)
945/30 = a1+a60
a1+a60 = 31.5
We have these two equations
a1+a60 = 31.5
a1+a40 = 21.5
Let's subtract the terms straight down
a1-a1 turns into 0 and goes away
a60-a40 is unknown
31.5-21.5 turns into 10
We can see that
a60-a40 = 10
which rearranges to
a60 = a40+10
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The n = 40th term is defined as follows
an = a1 + d(n-1)
a40 = a1 + d(40-1)
a40 = a1 + d(39)
a40 = a1 + 39d
Similarly, n = 60 leads to
an = a1 + d(n-1)
a60 = a1 + d(60-1)
a60 = a1 + d(59)
a60 = a1 + 59d
Now we'll use a60 = a40+10 to get
a60 = a40+10
a1 + 59d = a1 + 39d+10
59d = 39d+10 ............ the a1 terms cancel
59d-39d = 10
20d = 10
d = 10/20
d = 1/2
d = 0.5
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Recall that,
Sn = (n/2)*(a1+an)
We can replace an with a1+d(n-1) and simplify
Sn = (n/2)*(a1+an)
Sn = (n/2)*(a1+a1+d(n-1))
Sn = (n/2)*(2a1+d(n-1))
This is an alternative formula to sum the first n terms
We'll use this alternative formula to help determine a1
Let's plug in n = 40 and d = 0.5
Sn = (n/2)*(2a1+d(n-1))
S40 = (40/2)*(2a1+0.5(40-1))
430 = 20*(2a1+19.5)
430/20 = 2a1+19.5
21.5 = 2a1+19.5
21.5-19.5 = 2a1
2 = 2a1
2a1 = 2
a1 = 2/2
a1 = 1
Or we could use n = 60 with d = 0.5
Sn = (n/2)*(2a1+d(n-1))
S60 = (60/2)*(2a1+0.5(60-1))
945 = 30*(2a1+29.5)
945/30 = 2a1+29.5
31.5 = 2a1+29.5
31.5-29.5 = 2a1
2 = 2a1
2a1 = 2
a1 = 2/2
a1 = 1
We end up with the same starting term.
The last thing to do is to plug a1 = 1 and d = 0.5 into the nth term formula below
an = a1 + d(n-1)
an = 1 + 0.5(n-1)
an = 1 + 0.5n-0.5
an = 0.5n + 1-0.5
an = 0.5n + 0.5
an = 0.5(n + 1)
an = (1/2)(n + 1)
an = (n + 1)/2
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As a way to check, plug in n = 40
an = (n + 1)/2
a40 = (40 + 1)/2
a40 = 20.5
Then note how
Sn = (n/2)*(a1+an)
S40 = (40/2)*(1+20.5)
S40 = 430
which matches with the instructions
Also, plug in n = 60
an = (n + 1)/2
a60 = (60 + 1)/2
a60 = 30.5
which means,
Sn = (n/2)*(a1+an)
S60 = (60/2)*(1+30.5)
S60 = 945
and that matches as well. The answer has been confirmed.
A longer winded way of checking the answer is to list the first 40 terms of the sequence (which is now possible since we know a1 = 1 and d = 0.5), then add up said terms. Do the same for the first 60 terms as well. This is not recommended unless you are using computer software to make the process relatively painless.
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Answer: (n + 1)/2
Answer by muthbab@gmail.com(5) (Show Source):
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