To solve this problem, we need to find consecutive positive integers summing up to 210 such that the number of integers in the sequence is as large as possible. Let’s work through this step-by-step.
1. Identify the Sum Formula for Consecutive Integers:
The sum of the first \( n \) consecutive positive integers starting from 1 is given by the formula:
[tex]\[
S = \frac{n(n + 1)}{2}
\][/tex]
where \( S \) is the sum and \( n \) is the number of terms.
2. Determine \( n \):
We want to find the largest \( n \) such that the sum of these consecutive integers equals 210. We start incrementing \( n \) and compute the sum until it reaches or exceeds 210.
3. Check When Sum Equals 210:
a. Start with \( n = 1 \) and keep summing until the total just exceeds or equals 210.
[tex]\[ \frac{1(1 + 1)}{2} = 1 \][/tex]
[tex]\[ \frac{2(2 + 1)}{2} = 3 \][/tex]
[tex]\[ \frac{3(3 + 1)}{2} = 6 \][/tex]
[tex]\[ \frac{4(4 + 1)}{2} = 10 \][/tex]
[tex]\[ \frac{5(5 + 1)}{2} = 15 \][/tex]
[tex]\[ \frac{6(6 + 1)}{2} = 21 \][/tex]
[tex]\[ \frac{7(7 + 1)}{2} = 28 \][/tex]
[tex]\[ \frac{8(8 + 1)}{2} = 36 \][/tex]
[tex]\[ \frac{9(9 + 1)}{2} = 45 \][/tex]
[tex]\[ \frac{10(10 + 1)}{2} = 55 \][/tex]
[tex]\[ \frac{11(11 + 1)}{2} = 66 \][/tex]
[tex]\[ \frac{12(12 + 1)}{2} = 78 \][/tex]
[tex]\[ \frac{13(13 + 1)}{2} = 91 \][/tex]
[tex]\[ \frac{14(14 + 1)}{2} = 105 \][/tex]
[tex]\[ \frac{15(15 + 1)}{2} = 120 \][/tex]
[tex]\[ \frac{16(16 + 1)}{2} = 136 \][/tex]
[tex]\[ \frac{17(17 + 1)}{2} = 153 \][/tex]
[tex]\[ \frac{18(18 + 1)}{2} = 171 \][/tex]
[tex]\[ \frac{19(19 + 1)}{2} = 190 \][/tex]
[tex]\[ \frac{20(20 + 1)}{2} = 210 \][/tex]
When \( n = 20 \), the sum equals exactly 210.
4. Conclusion:
The largest integer in this sequence, when 210 is written as the sum of the largest possible number of consecutive positive integers, is the final integer in the sequence, \( 20 \).
Thus, when 210 is expressed as the sum of the greatest possible number of consecutive integers, the largest integer in the sequence is [tex]\( 20 \)[/tex].
