consecutive terms in sequence

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Nekkamath

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Many numbers ca be made by adding two or more consecutive terms of the arithmetic sequence 2,5,8,11,... Two such examples are 7=2+5 and 24=5+8+11. What is the smallest number that can be made in at least two different ways by adding consecutive terms of this sequence?

Do you have to add others number to the sequence, such as 14,17,20, 23... to get the answer?

2+5+8=15, 2+5+8+11=26, 11+14=26
 
Nekkamath said:
Many numbers ca be made by adding two or more consecutive terms of the arithmetic sequence 2,5,8,11,... Two such examples are 7=2+5 and 24=5+8+11. What is the smallest number that can be made in at least two different ways by adding consecutive terms of this sequence?

Do you have to add others number to the sequence, such as 14,17,20, 23... to get the answer?

2+5+8=15, 2+5+8+11=26, 11+14=26<-- that should be 25
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Many numbers ca be made by adding two or more consecutive terms of the arithmetic sequence 2,5,8,11,... Two such examples are 7=2+5 and 24=5+8+11. What is the smallest number that can be made in at least two different ways by adding consecutive terms of this sequence?

Consider the following sequence:

n....1......2......3......4......5......6......7......8......9......10......11......12......13......14......15
N....2.....5......8.....11.....14....17....20....23....26......29......32......35......38......41......44

This is an arithmetic progression with first term a = 2 and common difference d = 3. (Note the repeat of the units place digit)

The nth term of this arithmetic sequence is N = a + (n - 1)d

The sum of n terms from n = 1 to n = N is S = n(a + N)/2 = n(a + a + (n - 1)d)/2 = (3n^2 + n(2a - 3))/2

What is the smallest sum of terms that can be represented by two different sums of consecutive terms of the sequence?

One of the sums is logically one starting with the first term 2. The other logically one of two higher terms if the sum is too be a minimum.

S = the sum of successive terms from 2 on up = (3n^2 + n)/2 (a = 2 making 2a - 3 = 1)

The sum of two successive higher terms derives from S' = (a + (n' - 1)d) + (a + (n' - 1)d + 3) = 2(a + (n' - 1)d + 3

For a = 2 and d = 3, S' = 2(2) + 2(n' - 1)3 + 3 = 6(n' - 1) + 7.

Since the two sums must be equal, 6(n' - 1) + 7 = (3n^2 + n)/2 from which 3n^2 + n - (12(n' - 1) + 14 = 0

Then, n = [-1+/-sqrt(1^2 + 12(12(n' - 1) + 14))]/6 or n = [-1+/-sqrt(144n' + 25)]/6

What values of n' will make 144n' + 25 a perfect square? Remember, n' must be larger than n.

The first value of n' that meets our criteria is n' = 31 which leads to an n = 11.

These values yield the sum of the first 11 terms as

2 + 5 + 8 + 11 + 14 + 17 + 20 + 23 + 26 + 29 + 32 = 187 or

S = (3(11)^2 + 11)/2 = 187 and the sum of the 31st and 32nd terms as

S' = 92 + 95 = 187.

I believe this is the smallest sum that can be represented by two differents sums of successive terms of the sequence.

An n' = 50 and an n = 14 also produces equal sums of 301.

It is possible, by the same means, to derive a sum of the first "n" terms that is equal to a single larger term. An example is the sum of the first 4 terms equals the 9th term; 2 + 5 + 8 + 11 = 26.

Similarly, the sum of the first "n" terms is equal to 3 larger consecutive terms. An example is the sum of the first 12 terms equals the sum of the 24th + 25th + 26th terms ; 2 + 5 + 8 + 11 + 14 + 17 + 20 + 23 + 26 + 29 + 32 + 35 = 71 + 74 + 77 = 222.

I am sure there are other higher sums.
 
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