Set S consists of n consecutive odd integers. The sum of all elements in Set S is 195. If … — GMAT

Set $S$ consists of $n$ consecutive odd integers. The sum of all elements in Set $S$ is $195$ . If the largest element in Set $S$ is $9$ times the smallest element in Set $S$ , what is the value of $n$ ?

A
$3$
B
$9$
C
$12$
$13$ Cevap
E
$15$

Cevap

The total number of elements in Set

S

13

For any set of consecutive odd integers, the terms are spaced

2

units apart. Expressing the largest term as

a + 2(n-1) = 9a

yields

n = 4a + 1

. Using the average formula for evenly spaced sets,

\text{Sum} = n \times \frac{\text{first} + \text{last}}{2} = n(5a) = 195

, giving

an = 39

. Substituting

n = 4a + 1

gives

4a^2 + a - 39 = 0

, which factors as

(4a + 13)(a - 3) = 0

, so

a = 3

. Substituting

a = 3

into

n = 4a + 1

gives

n = 13

Adım Adım Çözüm

Express the largest term in terms of the smallest term

a

and term count

n

The largest term is

a + 2(n - 1)

. Since the largest term is

9a

, we have

a + 2(n - 1) = 9a \implies 2(n - 1) = 8a \implies n - 1 = 4a \implies n = 4a + 1

Consecutive odd integers increase by increments of

2

Express the sum of the set using the arithmetic mean of an evenly spaced set.

\text{Average} = \frac{\text{Smallest} + \text{Largest}}{2} = \frac{a + 9a}{2} = 5a. \text{Sum} = n \times \text{Average} \implies 195 = n(5a) \implies an = 39.

For any set of consecutive odd integers, the average is the mean of the first and last terms.

Substitute

n = 4a + 1

into

an = 39

and solve for

a

a(4a + 1) = 39 \implies 4a^2 + a - 39 = 0 \implies (4a + 13)(a - 3) = 0. Since

a

must be a positive integer,

a = 3

The smallest element of a set of positive odd integers must be a positive odd integer.

Calculate

n

using

a = 3

n = 4(3) + 1 = 13.

Substitute the value of

a

back into the formula derived in Step 1.

Anahtar Kavram

Consecutive Integers and Number Sets

Soru

Cevap

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