Question

13. The sum of three consecutive integers is 30 then the least number is​

Answer 1

Let three consecutive integers be (a) ,(a + 1),(a + 2)

$(a )+(a + 1 )+ (a + 2) = 30$

$\: \: \: \: \: \: \: \: \: \: \: \: \: 3a + 3 = 30$

$\: \: \: \: \: \: \: \: \: \: \: 3(a + 1) = 30$

$\: \: \: \: \: \: \: \: \: \: \: \: \: a \: + 1 \: = 10$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: a \: = 9$

$\red{a \: = 9}$

$\green{(a \: + 1) = (9 + 1) = 10}$

$\purple{(a + 2) =( 9 + 2) = 11}$

hence smallest number is 9.

$\sf\orange{please\:mark\:as\:brainliest............}$

Answer 2

Given:

Sum of three consecutive integers = 30

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To find:

The least number.

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Solution:

• Let the first number be x.
• Let the second number be (x+1).
• Let the third number be (x+2).

Now we are given that sum of all the three numbers is 30.

So,

x + (x+1) + (x+2) = 30

$\implies \sf {x+x+1+x+2 = 30}$

$\bigstar {\sf {\orange {Taking\ the\ variables\ on\ one\ side\ and\ constants\ on\ the\ other.}}}$

$\implies \sf {x+x+x+1+2 = 30}$

$\implies \sf {3x+3 = 30}$

$\implies \sf {3x = 30-3}$

$\implies \sf {3x = 27}$

$\implies \sf {x = \dfrac {27}{3}}$

$\boxed {\bf {\purple {x = 9}}}$

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Verification:

On substituting the value of x as 9 in the equation,

x + (x+1) + (x+2) = 30

$\implies \sf {9+(9+1)+(9+2) = 30}$

$\implies \sf {9+10+11 = 30}$

$\implies \sf {30 = 30}$

LHS = RHS

Hence Verified!

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The numbers are:

• First number = x

= 9

• Second number = x+1

= 9+1

= 10

• Third number = x+2

= 9+2

= 11

The least number is 9.