Question

If sum of 4 consecutive natural odd numbers is 32 then find smallest number​

Answer 1

Step-by-step explanation:

Assume that the first number is x

Then the sum is:

x+ (x+2) + (x+4) + (x+6) =-32

4x + 12 = -32

4x = -32 -12 = -44

==> x= -11

Then the numbers are:

-11, -9, -7 , and -5 ( All odd, negative, and consecutive)

To check:

-11 + -9 + -7 + -5 = -32

Answer 2

Step-by-step explanation:

Given :

Sum of 4 consecutive natural odd numbers = 32

To find :

Smallest number.

Solution :

Let the number be x

Here,

The four natural odd numbers with variables are :

x

x + 2

x + 4

x + 6

Sum of these = 32

Equation will be :-

x + x + 2 + x + 4 + x + 6 = 32

Calculations :-

x + x + 2 + x + 4 + x + 6 = 32 = 4x + 12=32 = 32 - 12 = 4x = 20 / x = 5

⟹x+x+2+x+4+x+6=32

⟹4x+12=32

⟹4x=32−12

⟹4x=20

⟹x=

4

20

⟹x=5

Therefore, The smallest number is 5.

Other odd natural number are :-

x + 2 = 5 + 2 = 7

x + 4 = 5 + 4 = 9

x + 6 = 5 + 6 = 11

All 4 numbers are, 5,7, 9,11.

Verification :-

x + x + 2 + x + 4 + x + 6 = 32 = 5 + 5 + 2 + 5 + 4 + 5 + 6 = 32 = 32 = 32 LHS = RHS Hence, Verified.

⟹x+x+2+x+4+x+6=32

⟹5+5+2+5+4+5+6=32

⟹32=32

LHS=RHS

Hence,Verified.