Question

Which of the following could be the sum of any 12 consecutive natural numbers?

A 92

B 198

C 328

D 412

E 1,570

Answer 1

the answer is 1570 is the sum of 12 consecutive question

Answer 2

Answer:

Option (B) 198 cab be written as the sum of any 12 consecutive natural numbers.

Step-by-step explanation:

Let 12 consecutive natural numbers be $x,(x+1),(x+2),(x+3) ,$$(x+4),$$(x+5),(x+6),(x+7),(x+8),(x+9),(x+10),(x+11)$

According to the question,

Sum of any 12 consecutive natural is 198.

A. 92

⇒ $x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)+(x+7)$$+(x+8)+(x+9)+(x+10)+(x+11)$ $=92$

⇒ $12x+66=92$

⇒ $12x=92-66$

⇒ $x=26/12$

⇒ $x=13/6$, which is not possible

As $x$ is a natural number. Hence, it cannot be in fraction.

Thus, Option (A) is incorrect.

B. 198

⇒ $x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)+(x+7)$$+(x+8)+(x+9)+(x+10)+(x+11)$ $=198$

⇒ $12x+66=198$

⇒ $12x=198-66$

⇒ $x=132/12$

⇒ $x=11$, which is possible

Thus, Option (B) is correct.

C. 328

⇒ $x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)+(x+7)$$+(x+8)+(x+9)+(x+10)+(x+11)$ $=328$

⇒ $12x+66=328$

⇒ $12x=328-66$

⇒ $x=262/12$

⇒ $x=131/6$, which is not possible

As $x$ is a natural number. Hence, it cannot be in fraction.

Thus, Option (C) is incorrect.

D. 412

⇒ $x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)+(x+7)$$+(x+8)+(x+9)+(x+10)+(x+11)$ $=412$

⇒ $12x+66=412$

⇒ $12x=412-66$

⇒ $x=346/12$

⇒ $x=173/6$, which is not possible

As $x$ is a natural number. Hence, it cannot be in fraction.

Thus, Option (D) is incorrect.

E. 1570

⇒ $x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)+(x+7)$$+(x+8)+(x+9)+(x+10)+(x+11)$ $=1570$

⇒ $12x+66=1570$

⇒ $12x=1570-66$

⇒ $x=1504/12$

⇒ $x=376/3$, which is not possible

As $x$ is a natural number. Hence, it cannot be in fraction.

Thus, Option (E) is incorrect.

#SPJ3