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

s1 = 29(1)-1^2 = 28
hence a1 = 28
s2 = 29(2)-2^= 58-4 = 54
hence a2 = 54-28 = 26
s3= 29(3)-3^2 = 87-9 = 78
a3= 78-54= 24
hence each additional term will have the difference by 2
therefore n= 13

Using \(T_n= S_n- S_{n-1}\)
\(=> T_n= 20-2n=4\)
\(=> a_k=30-2k=4\)
\(=> k=\frac{26}{2}=13\)
Hence B)

sum of the sequence is given by \(S_n= 29n - n^2\)

s1 = 28
s2=54
s3=78
s4=100
common difference is 2
at s12 = 204
s13 = 208
difference is 4
13th term
OPTION B is correct

Answer is B.

sum of AP = (n/2)*(2a + (n-1)d) where d is common difference and a is first term.
using given, its n(29-n) = (n/2)*(2a + (n-1)d)
since n > 0 , 29-n = (1/2)*(2a + (n-1)d) = (a - d/2) + (d/2)n
solve d/2 = -1 and (a-d/2) = 29 to get d = -2, a = 28

given a-k = 4 . implies a + (k-1)d = 4. plus a and d values. u get k = 13. answer is B

Given that Sn = 29n - n^2
S1 = a1 = 29*1 -1^2 = 28
S2 = a1 + a2 = 29*2 - 2^2 = 54
So, a2 = 26
Therefore, d = -2
ak = a1 + (k-1)*d = 28 + (k-1)*-2 = 28 - 2k + 2
Now given that ak = 4, we have k = 13.

So, the answer is B.

s1=29(1)-1
=28
a1=28
s2=29(2)-4=54
a2=54-28=26
d=26-28=-2
ak=a1+(k-1)d
4=28+(k-1)-2
-24/-2=k-1
13=k ans

S1 will be 29(1)-(1)^2= 28 =a1

S2 will be 29(2)-(2)^2= 54
S2 = a1+a2

Putting value of a1 we get a2 =54-28 = 26

Hence d =a2-a1 =-2

Now ak term =4= a1+(k-1)*d

Solving by putting d=-2 and a1= 28

We will get k as 13

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\(S_n= 29n - n^2\)

\(a_1\) = \(S_1\) & \(a_2\) = \(S_2 - a_1\)

\(S_1= 29 * 1 - 1^2\)

\(a_1 = 28\)

Solving for n=2, we get
\(a_2 = 26\)

Now
\(a_k = 4\)
\(a_1 + (k-1)d = 4\)
\(28 + (k-1) * (-2) = 4\)
\(k = 13\)

Ans B

The answer is E- 100
ak=4 and the sum of k terms in the sequence comes out to be 100

ak = S(k) - S(k-1)

We know that S(k) = 29K - k^2. .......... eq1
S(k-1) = 29(k-1) - (k-1)^2
S(k-1) = 29k - 29 - (K^2 +1 - 2k)
S(k-1) = 29k -29 - k^2 -1 +2k ........... eq2

eq1-eq2
-> 29K - k^2 - (29k -29 - k^2 -1 +2k)
-> 29K - k^2 - 29k + 29 + k^2 +1 - 2k
-> 30-2k

we know that ak = 4
Therefore,
30-2k = 4
2k = 26
k=13

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.

The sum of the first n terms of an arithmetic progression, a1, a2, ..., is given by Sn=29n−n^2. If ak=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

Solution:
Given, sum of first n terms of an arithmetic progression is given by Sn=29n−n^2.
So, S1 will be the sum of only the first term or rather S1 will be the first term. Calculating S1=29(1)-(1)^1=28.
So, a1=28.
Now, S2 will be sum of first two terms. S2=29(2)-(2)^2=54.
So, a1 + a2 = 54. But a1=28, so a2=26.
Similarly from S3 we get a3= 24.
So a1=28, a2=26, a3=24 is an arithmetic progression with a common difference of -2.
Calculating kth term in such a progression, ak = a1 + (k-1)d............(1)
Given ak=4, we know d=-2. Substituting these values in the equation (1) we get, k=13. ANSWER B

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_n = 29N - N ^2 = N (29 - N)
for n = 1, sum = 28
so a1= 28

sum of two digits = 2(27)= 54
so a1+a2= 54
a2=26

sum of three digits = 3*26= 78
a1+a2+a3 = 78
28+26+a3 =78
a3=78- 54 = 24

so with every digit, 2 is decreased, so numbers will be
28, 26, 24, 22, 20, 18, 16, 14, 12, 10, 8, 6, 4

Ak = a13

k = 13

\(S_n= 29n - n^2\)

\(S_1=29-1=28 => a_1=28\)

\(S_2= 29*2 - 4= 54 => a_2=26\)[/m]

Since its an AP we now know
\(a_1=28\\
d=-2\\
\)
\(a_n=4\)
\(a + (n-1)d = 4\\
28 + (n-1)-2 = 4\\
n =13\)

Answer : B. 13

Put n = 1, S1 = 29-1 = 28 so first term = 28.
Put n = 2, S2 = 54 so second term = 54-28 = 26.
Common difference = -2.

kth term in A.P. = first term + (k-1)(common difference) = 28+(k-1)(-2)
put ak = 4, 28 - 2(k-1) = 4

k = 13.

IMO B.

sum of n terms of A.P = n/2 *[2a+(n-1)*d]=29n-n^2
since n=0, hence dividing both the side by n

a+(d/2)*n-d/2= 29n-n

comparing both the sides
d/2=-1, d=-2
a-d/2=29, a =28

now kth term of the A.P
kth term = a +(k-1)d
4= 28 +(k-1)* (-2)

k=13, hence B should be the answer.

Let c be the progressive constant of the sequence
a[n]=a1+(n-1)c

Sum = (a1+a[n])*n/2 = (2(a1) + (n-1)c)*n/2 = n*a1 + n(n-1)*c/2 (1)
29n-n^2= 28n - n(n-1) (2)

From (1)=(2), we get a1=28, c=-2

a[k]=a1 + (k-1)c = 28 + (k-1)(-2)=4 -> k=13

Answer B