Question

The sum of 28 consecutive whole numbers is 686. Find the smallest addend in the sum.

Answer 1

The smallest number in the sum of 28 consecutive numbers totaling 686 is 11. This is found by calculating the average of the sequence (686 ÷ 28 = 24.5) and working backwards to the first number.

Step-by-step explanation:

To find the smallest number in the sum of 28 consecutive whole numbers that equals 686, let's denote the smallest number as n. Because the numbers are consecutive, the next number would be n+1, then n+2, and so on, up to n+27, the largest number in the sequence. The average of these numbers is the middle number, which would be the average of the 14th and 15th terms, n+13 and n+14. Since the average of the sequence is the sum divided by the number of terms (ø6ø6∘28), we can find the middle of the sequence, then calculate back to find the smallest number.

The average of the sequence is 686 ÷ 28 = 24.5. The middle numbers are the 14th and 15th terms, so n+13 and n+14. Since the average of these two numbers is 24.5, we can set up the equation n + 13.5 = 24.5. Solving for n gives us n = 24.5 - 13.5 = 11. Therefore, the smallest addend in the sum is 11.

Answer 2

`\bf \textit{28 consecutive numbers}\to \begin{cases} a\\ a+1\\ a+2\\ a+3\\ a+4\\ a+5\\ a+6\\ a+7\\ a+8\\ a+9\\ a+10\\ a+11\\ a+12\\ a+13\\ a+14\\ a+15\\ a+16\\ a+17\\ a+18\\ a+19\\ a+20\\ a+21\\ a+22\\ a+23\\ a+24\\ a+25\\ a+26\\ a+27 \end{cases}`

now, their sum is 686
meaning

(a)+(a+1)+(a+2)+(a+3)+(a+4)+(a+5)+(a+6)+(a+7)+(a+8)+(a+9)+(a+10)+(a+11)+(a+12)+(a+13)+(a+14)+(a+15)+(a+16)+(a+17)+(a+18)+(a+19)+(a+20)+(a+21)+(a+22)+(a+23)+(a+24)+(a+25)+(a+26)+(a+27) = 686

solve for "a"