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it can be easily solved with a tad bit of guesswork.

we know, 1+2+3=6, and 3+4=7. since it's seven consecutive integers

so, that leaves us with only this feasible option -- -2, -1, 0, 1, 2, 3, 4.

mean of smallest & largest numbers -> (-2+4)/2 = 1
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The sum of seven consecutive integers is equal to 7. What is the average of the greatest and least terms in the sequence?

Let consecutive integers= a, a+1, a+2, a+3, a+4, a+5, a+6
Sum of these integers= 7
7a+21=7
a=-2

Integers are -2,-1,0,1,2,3,4

Average of least and greatest= -2+4/2=2/2=1

D
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7 consecutive integers: n ,n+1...n+6

Acc to ques.
7n+21=7
N=-2
Largest number -2+6=4
average=4+(-2)=2/2=1

Option D
Bunuel
The sum of seven consecutive integers is equal to 7. What is the average of the greatest and least terms in the sequence?

(A) 7
(B) 4
(C) 2
(D) 1
(E) -2


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We call \(a\) be the first term, representing "seven consecutive integers" as \(a,\ a+1,\ a+2,\ a+3,\ a+4,\ a+5,\ a+6\).
\(a,\ a+1,\ a+2,\ a+3,\ a+4,\ a+5,\ a+6\)

We translate "The sum of seven consecutive integers is equal to 7" into \(a + (a+1) + (a+2) + (a+3) + (a+4) + (a+5) + (a+6) = 7\).

This can be simplified to
\(7a + 21 = 7\)

We solve the resulting linear equation for the unknown.
\(7a = -14\)
\(a = -2\)

We identify the least and greatest terms by substituting \(a\) into the sequence.
\(\text{least term} = a = -2\) \(\text{greatest term} = a + 6 = 4\)

We use the average-of-first-and-last formula for evenly spaced integers to find the average of the least and greatest terms.
\(\frac{-2 + 4}{2} = 1\)

Answer 1

Hope this helps!
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