A set S consists of k consecutive integers. The sum of the first m integers in set S is 34… — GMAT

A set $S$ consists of $k$ consecutive integers. The sum of the first $m$ integers in set $S$ is $34$ , and the sum of the last $m$ integers in set $S$ is $74$ . If the sum of all $k$ integers in set $S$ is $189$ , what is the value of $k$ ?

Cevap: 14

Cevap

The total number of integers in set S is 14.

By applying the property that the arithmetic mean of an evenly spaced set is the average of the mean of its first

m

elements and the mean of its last

m

elements, we establish that the mean of the set is

\frac{54}{m}

. Since the total sum is

189

k \cdot \frac{54}{m} = 189

, which yields

k = 3.5m

. Substituting this ratio into the difference between the two subset sums

m(k-m) = 40

gives

2.5m^2 = 40

, so

m = 4

and

k = 14

Adım Adım Çözüm

Set up algebraic expressions for the sums of the first m terms and last m terms.

Let the set be

S = \{a, a+1, \dots, a+k-1\}

. The first

m

terms sum to

S_1 = m a + \frac{m(m-1)}{2} = 34

. The last

m

terms sum to

S_2 = m(a+k-m) + \frac{m(m-1)}{2} = 74

Consecutive integer sums can be represented by the starting term and the number of terms.

Subtract the sum of the first m terms from the sum of the last m terms.

S_2 - S_1 = m(a+k-m) - ma = m(k-m) = 74 - 34 = 40

Subtracting eliminates the initial term

a

and quadratic term

\frac{m(m-1)}{2}

, giving a clean relationship between

m

and

k

Determine the arithmetic mean of set S using subset averages.

The average of the first

m

terms is

\frac{34}{m}

and the average of the last

m

terms is

\frac{74}{m}

. The average of the entire set is the midpoint of these two averages:

\text{Mean} = \frac{1}{2}\left(\frac{34}{m} + \frac{74}{m}\right) = \frac{54}{m}

In any evenly spaced set, the overall median/mean is equal to the average of the lower-bound subset mean and upper-bound subset mean.

Relate the total sum to the set size k and overall mean.

S_{total} = k \times \text{Mean} \implies 189 = k \left(\frac{54}{m}\right) \implies \frac{k}{m} = \frac{189}{54} = 3.5 \implies k = 3.5m

The sum of a set of consecutive integers is always equal to the number of terms times the mean of the set.

Solve for m and k using the system of equations.

Substitute

k = 3.5m

into

m(k-m) = 40 \implies m(2.5m) = 40 \implies 2.5m^2 = 40 \implies m^2 = 16 \implies m = 4

. Thus,

k = 3.5 \times 4 = 14

Since

m > 0

, taking the positive square root gives

m = 4

, which leads directly to

k = 14

Anahtar Kavram

Average and Sum Equivalences in Evenly Spaced Sets

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