At the middle of a sequence!
An arithmetic sequence has 11 terms which sum to 220. What is the middle (6th) term in the sequence?
The answer is 20.
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8 solutions
GRATE DUDE
Let the middle term be x .
Then the sequence will be : ( x − 5 d , … , x − d , x , x + d , … , x + 5 d ) , in order to have just 11 terms , where d is a constant and the common difference in that sequence .
Therefore, the sum of terms , which is given by 2 2 0 , will be : + ( x − 5 d ) ( x + 5 d ) + + … … + + ( x − d ) ( x + d ) x 1 1 x
Hence, 1 1 x = 2 2 0 x = 2 0
Therefore, the middle term is 2 0
Moderator note:
Prasun Biswas is right. More generally, the problem did not explicitly say that all the terms are integers, nor did it say that it is non constant. So " d ∈ Z − {0}" is not applicable.
There's a much more simpler approach to this problem.
Here's how I did it. Total number of terms is n = 1 1 , denote first term as a and common difference as d , apply the arithmetic progression sum formula: 2 n ( 2 a + ( n − 1 ) d ) = 2 2 0 , thus 2 1 1 ( 2 a + 1 0 d ) = 2 2 0 or a + 5 d = 2 0 , equivalently a + ( 6 − 1 ) d = 2 0 which is the 6th term.
Another formula for the sum of the first n terms of an A.P. is 2 n ( a 1 + a n ) , where a n is the n th term.
Are you sure that d ∈ Z − { 0 } ?
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Thank you for your useful note , I have modified the solution so that d ∈ R − {0} instead of d ∈ Z − {0} .
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Why restrict d ? By definition of common difference of an arithmetic progression, the common difference can be any constant, not restricted to non-zero reals. Also, you can even have d = 0 or even complex non-real value for d . There is absolutely no need to restrict the domain of d .
As an example, consider the following arithmetic progression:
{ 2 0 − 5 i , 2 0 − 4 i , … , 2 0 + 4 i , 2 0 + 5 i } where i = − 1 ∈ / R
This is also a valid arithmetic progression with common difference i and the sum of all the terms is 2 2 0 with the middle term being 2 0 .
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@Prasun Biswas – Thank you once more for your concern and guiding . But if we took d (the common difference) = 0 , there would be no progression at all . And this is why d must be not equal 0 . And as for imaginary numbers to be terms of an arithmetic sequence , I think this is away from what AP deals with , but you are right at the end .
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@Mohamed Ahmed Abd El-Fattah – I'm getting what you mean but you should know that, mathematically, if we have d = 0 , that would be termed as an arithmetic progression too. The definition of arithmetic progression is "a sequence of values such that the difference between two consecutive values of the sequence is constant".
There is little relevance of the literal meaning of the word "progression" in mathematical context. There is no reason to restrict d for an arithmetic progression in general. Even for the problem here, we need not restrict d because the number of terms of the AP is odd. In my opinion, you should remove the part where you restrict d and replace it with the following:
d is a constant (common difference for the AP here).
Even if you see the Wikipedia article on AP, you can see that they simply mention d to be a constant and not impose any restrictions on it.
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@Prasun Biswas – OK, I got it , and I have done as what you said , but could you please explain what you meant by : "we need not restrict d because the number of terms of the AP is odd." .
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@Mohamed Ahmed Abd El-Fattah – See my first comment on the solution by Kevin Li.
The first term is x . Therefore, the second term is x + y and the third term is x + 2 y . x + x + 2 y + x + 3 y + . . . + x + 1 0 y = 1 1 x + 5 5 y . 1 1 x + 5 5 y = 2 2 0 x + 5 y = 2 0 Since x + 5 y is the sixth term, which is the middle term, the middle term is equal to 20.
Moderator note:
Although you're right, it's impractical to calculate the value of 1 + 2 + … 1 0 or 1 + 2 + … 1 1 especially when the numbers get larger. And yes, you might be tempted to mention that m = 1 ∑ n m = 2 n ( n + 1 ) . Nonetheless Good work.
Bonus question: What would the answer be if I replace the number 11 by 999 and the number 220 by 10989?
[Response to Challenge Master Note]
We can do a lot better than that. If we replace 2 2 0 by any value x ∈ C and 1 1 by any odd positive integer m = 2 k + 1 , k ∈ Z 0 + , then we can formulate, w.l.o.g, the following AP sequence: { x + j δ } j = − k j = k where δ ∈ C is a constant (common difference for the AP). It is obvious that x is the middle term of this AP. The sum is,
S = j = − k ∑ k ( x + j δ ) = ⎝ ⎛ j = − k ∑ k x ⎠ ⎞ + δ ⋅ ⎝ ⎛ j = − k ∑ k j ⎠ ⎞ = ( 2 k + 1 ) x + δ × 0 = m x
For the problem in the note, it's given that S = 1 0 9 8 9 and m = 9 9 9 (odd) and we want the middle term x . So, we have here,
S = m x ⟹ 1 0 9 8 9 = 9 9 9 x ⟹ x = 9 9 9 1 0 9 8 9 = 1 1
Similarly, for the original problem, we have S = 2 2 0 and m = 1 1 (odd), so, we can easily evaluate the answer to be 2 0 .
Note that :
a+b+c+d+e+...+k = 220
Since it is arithmetic sequence, we can say that:
220 = a + (a+1) + (a+2) + ... (a+10)
220 = 11a + 55
There are 11a since there are 11 numbers, and 55 is the summation of numbers from 1 to 10
So this implies that :
a = (220-55) / 11
a = 15
Since we are looking for the 6th term then
6th term = a+5 = 15 + 5 = 20
Let there be an arithmetic progression with x terms in it, where x is an odd whole number. Let the difference between every consecutive term in the sequence be y and the first term be a
Hence, the exact middle term in the sequence would be the [(x-1)/2 + 1]th term. Now the value for that number in terms of a and y would be =
[a+ {(x-1)/2}+1-1)y
= [a+(x-1)/2 y ]
Now the sum of all the x numbers in the progression=
a+ [a+y] + [a+2y] + [a+3y] +......[a+(x-1)y]
= a x + [1+2+3+....+(x-1)] y
= a x + [ (x-1) x/2]*y
=x[a+(x-1)/2*y]
=x*the middle term
Hence the middle term= sum of the digits/total number of digits (for every finite arithmetic progression that has a total of odd numbers in itself)
Here x=11 and the sum is 220. Hence the middle term= 220/11=20
Note:The few concepts I have used here are-
1) the n th term of any Arithmetic progression can be expressed as
a+ (n-1)y , where a is the first number and y is the difference between successive numbers
2) The sum of 1+2+3+4+5...+n= n*(n+1)/2
PS: I have used * to denote multiplication
Moderator note:
To find the average in an arithmetic progression, we do not need to find the sum of all of these numbers. We just need to realize that each pair of end terms have an average that is the middle terms.
Really simple.
We know that: Sn= n/2 (2a+(n-1) d)...eqn1
Where Sn denotes sum of n terms with first term being "a" and common difference of the AP: d. Equating eqn1 with 220; we get:
20=a+5d Great! When we compare this with the general expression of nth term of an AP (being, Tn= a+(n-1) d) we see that thus is actually the 6 term. And fortunately, we were supposed to find the sixth term so 20 is the answer. It didn't require much logic tho, just formula application.
m i d d l e t e r m = n u m b e r o f t e r m s s u m = 1 1 2 2 0 = 2 0
easy. 1 + 2 + 3 + 4 + . . . . . . . . + n = 100 × 101.
let x be 1. we can have j as 2. thus. x = j.
when the lim as x > x^2. we could add 11x + 55y to the formula.
1 s t T e r m = A + D 6 t h T e r m = A + 6 D = X 1 1 t h T e r m = A + 1 1 D 2 2 0 = 1 1 A + ( 1 1 t h T r i a n g l e ) D 2 2 0 = 1 1 A + ( 1 1 ) ( 1 2 ) ( 0 . 5 ) D 2 2 0 = 1 1 A + 6 6 D A + 6 D 1 1 A + 6 6 D = 1 1 1 1 = X 2 2 0 1 1 2 2 0 = X 2 0 = X
To get the sum of an arithmetic sequence,
sum = average x number of terms
We can use the formula above to get the middle term, or simply the average of the sequence.
Hence,
denote middle term as n
220 = n x 11
220 = 11n
220/11 = 11n/11
20 = n