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Around the World in 80 Questions (Day 9): The sum of the first n term

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Bunuel
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Around the World in 80 Questions (Day 9): The sum of the first n term [#permalink] Around the World in 80 Questions (Day 9): The sum of the first n term   27 Jul 2023, 08:00
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The sum of the first n terms of an arithmetic progression, \(a_1, \ a_2, \ ...\), is given by \(S_n= 29n - n^2\). If \(a_k = 4\), what is the value of k ? (An arithmetic progression is a sequence of numbers such that the difference between the consecutive terms is constant)

A. 12
B. 13
C. 14
D. 15
E. 100


 


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Re: Around the World in 80 Questions (Day 9): The sum of the first n term [#permalink] Around the World in 80 Questions (Day 9): The sum of the first n term   Updated on: 28 Jul 2023, 08:09
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Given: The sum of the first n terms of an arithmetic progression, \(a_1, \ a_2, \ ...\), is given by \(S_n= 29n - n^2\).
Asked: If \(a_k = 4\), what is the value of k ? (An arithmetic progression is a sequence of numbers such that the difference between the consecutive terms is constant)

\(S_n = 29n - n^2 \)
\(S_{n-1} = 29(n-1) - (n-1)^2\)

nth term a_n is difference between S_n & S_{n-1} since only a_n is added to S_{n-1} to get S_n

\(a_n = S_n - S_{n-1} = 29n - n^2 - {29(n-1) - (n-1)^2} = 29 - (2n-1) = 30 - 2n\)

\(a_k = 30 - 2k = 4\)
26 = 2k
k = 13

IMO B

Originally posted by Kinshook on 27 Jul 2023, 08:22.
Last edited by Kinshook on 28 Jul 2023, 08:09, edited 1 time in total.
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Re: Around the World in 80 Questions (Day 9): The sum of the first n term [#permalink] Around the World in 80 Questions (Day 9): The sum of the first n term   27 Jul 2023, 08:39
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My Answer: B

a1=S1=28
S2=a1+a2=54 so, a2=26
d=a2-a1=-2

Now, ak=4=a1+(n-1)d
=28+(k-1)*-2
k=13
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Re: Around the World in 80 Questions (Day 9): The sum of the first n term [#permalink] Around the World in 80 Questions (Day 9): The sum of the first n term   27 Jul 2023, 09:43
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Bunuel wrote:
The sum of the first n terms of an arithmetic progression, \(a_1, \ a_2, \ ...\), is given by \(S_n= 29n - n^2\). If \(a_k = 4\), what is the value of k ? (An arithmetic progression is a sequence of numbers such that the difference between the consecutive terms is constant)

A. 12
B. 13
C. 14
D. 15
E. 100


 


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Good question as always.
\(S_n= 29n - n^2\)
We can find the first term by putting n=1
\(a_1=29*1-1^2=28\)
\(S_2=29*2-2^2=54\). Hence, \(a_2=54-28=26\)
Hence, commond difference=26-28=-2
Therefore, the AP is: 28, 26, 24...
to find the term of 4:
\(4=28+(n-1)*(-2)\)
\(n=13.\)
Hence B is correct.
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Re: Around the World in 80 Questions (Day 9): The sum of the first n term [#permalink] Around the World in 80 Questions (Day 9): The sum of the first n term   27 Jul 2023, 11:41
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Bunuel wrote:
The sum of the first n terms of an arithmetic progression, \(a_1, \ a_2, \ ...\), is given by \(S_n= 29n - n^2\). If \(a_k = 4\), what is the value of k ? (An arithmetic progression is a sequence of numbers such that the difference between the consecutive terms is constant)

A. 12
B. 13
C. 14
D. 15
E. 100


 


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for the Around the World in 80 Questions

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S1 = First Term = T1

S1 = 29 - 1 = T1 = 28

S2 = 29*2 - 4 = 58 - 4 = 54

T2 = S2 - S1 = 26

Difference = d = -2

an = T1 + (n-1)d

4 = 28 + (n - 1) * (-2)

-24 = (n - 1)* (-2)

12 = n - 1

n = 12+1 = 13

IMO B
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Re: Around the World in 80 Questions (Day 9): The sum of the first n term [#permalink] Around the World in 80 Questions (Day 9): The sum of the first n term   27 Jul 2023, 12:20
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Bunuel wrote:
The sum of the first n terms of an arithmetic progression, \(a_1, \ a_2, \ ...\), is given by \(S_n= 29n - n^2\). If \(a_k = 4\), what is the value of k ? (An arithmetic progression is a sequence of numbers such that the difference between the consecutive terms is constant)

A. 12
B. 13
C. 14
D. 15
E. 100


\(S_1\) = \(a_1\) = (29*1) - \(1^2\) = 29 - 1 = 28
\(S_2\) = \(a_1\) + \(a_2\) = (29*2) - \(2^2\) = 58 - 4 = 54
Therefore, \(S_2 = S_1 + a_2\)
=>\( 54 = 28 + a_2\)
=> \(54 - 28 = a_2\)
=>\( 26 = a_2\)
Since, the sequence is A.P.
then difference between two terms,d = a_2 - a_1 = 26 - 28 = -2

Substituting the values of \(a_n\) = \(a_k \)= 4, d and \(a_1 \)in the formula \(a_n = a_1 + (n-1)d\), we get
4 = 28 + (k-1)*(-2)
=> -24 = -2k + 2
=> 26 = 2k
=> 13 = k

Hence, OA should be B.
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Re: Around the World in 80 Questions (Day 9): The sum of the first n term [#permalink] Around the World in 80 Questions (Day 9): The sum of the first n term   27 Jul 2023, 18:59
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Bunuel wrote:
The sum of the first n terms of an arithmetic progression, \(a_1, \ a_2, \ ...\), is given by \(S_n= 29n - n^2\). If \(a_k = 4\), what is the value of k ? (An arithmetic progression is a sequence of numbers such that the difference between the consecutive terms is constant)

A. 12
B. 13
C. 14
D. 15
E. 100
 


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for the Around the World in 80 Questions

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Sum of the first n terms of an AP = (n/2)[2a + (n-1)d] where a is the first term and d is the common difference
Thus: (n/2)[2a + (n-1)d] = n(29 - n)
=> 2a + (n-1)d = 58 - 2n
=> (2a - d) + nd = 58 - 2n
Comparing: d = -2; 2a - d = 58 => a = (58 + d)/2 = 28

Thus, if the k-th term is 4: 28 + (k - 1)(-2) = 4
=> 28 - 2(k - 1) = 4
=> k - 1 = 12
=> k = 13

Answer B
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Re: Around the World in 80 Questions (Day 9): The sum of the first n term [#permalink]
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