Set A consists of k consecutive odd integers, and Set B consists of k consecutive even int… — GMAT

Set $A$ consists of $k$ consecutive odd integers, and Set $B$ consists of $k$ consecutive even integers, where $k > 1$ . The smallest integer in Set $B$ is $3$ greater than the median of Set $A$ . If all integers in Set $A$ are positive, the sum of all integers in Set $A$ is $145$ , and the median of Set $B$ is $36$ , what is the smallest integer in Set $A$ ?

A
21
B
23
25Answer
D
27
E
29

Answer

The smallest integer in Set A is 25.

For an arithmetic sequence of

k

consecutive odd integers, the mean and median are equal to

\frac{\text{Sum}}{k} = \frac{145}{k}

. The smallest integer in Set

B

b_1 = \frac{145}{k} + 3

. Since Set

B

consists of

k

consecutive even integers, its median is

b_1 + k - 1

. Equating this to

36

yields

\frac{145}{k} + k + 2 = 36

, leading to

k^2 - 34k + 145 = 0

, whose roots are

k=5

and

k=29

. If

k=29

, the median of Set

A

5

, which implies negative integers exist in Set

A

. Since all integers in Set

A

are positive,

k=5

. With

k=5

, the median of Set

A

29

, and the smallest integer is

29 - 2(2) = 25

Step-by-Step Solution

Express the median of Set A in terms of k.

M_A = \frac{145}{k}

For any evenly spaced set with an odd number of terms or symmetry, the arithmetic mean equals the median. The mean is the total sum divided by the number of terms

k

Express the median of Set B in terms of k using the given relationship for the smallest element of Set B.

M_B = \left(\frac{145}{k} + 3\right) + (k - 1) = \frac{145}{k} + k + 2

The smallest element in Set

B

b_1 = M_A + 3 = \frac{145}{k} + 3

. Since Set

B

contains

k

consecutive even integers (spacing

d=2

), its median is

b_1 + \frac{2(k-1)}{2} = b_1 + k - 1

Set the median of Set B to 36 and solve the quadratic equation for k.

k = 5

k = 29

Setting

\frac{145}{k} + k + 2 = 36

gives

\frac{145}{k} + k = 34

, which rearranges to

k^2 - 34k + 145 = 0

. Factoring gives

(k-5)(k-29) = 0

Determine the valid value of k and find the smallest integer in Set A.

Smallest integer in Set A is 25.

k = 29

M_A = \frac{145}{29} = 5

, and the smallest integer in Set

A

would be

5 - 2(14) = -23

, violating the condition that all integers in Set

A

are positive. Thus

k = 5

, making

M_A = 29

. The 5 consecutive odd integers are

25, 27, 29, 31, 33

, so the smallest integer is

25

Key Concept

Properties of consecutive integer sets, median-mean equivalence in arithmetic sequences, and term indexing.

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