Question

the sum of three consecutive multiples of 9 is 81 the least of the is

Answer 1

Proper question :- The sum of three consecutive multiples of 9 is 81. Which is the least among these three multiples?


Solution :-

Let the three consecutive multiples of 9 be x, x + 9 and x + 18

Their sum is 81


According to the given condition,

x + x + 9 + x + 18 = 81

➾ 3x + 27 = 81

➾ 3x = 81 - 27

➾ 3x = 54

➾ x = 18



∴ Smaller number ➾ x

➾ $\green{\boxed{\green{\boxed{\red{\textsf{18}}}}}}$


∴ Second number ➾ x + 9

➾ 18 + 9

➾ $\green{\boxed{\green{\boxed{\red{\textsf{27}}}}}}$


∴ Third number ➾ x + 18

➾ 18 + 18

➾ $\green{\boxed{\green{\boxed{\red{\textsf{36}}}}}}$

Answer 2

$\green{\boxed{\green{\boxed{\pink{\textsf{Solution:-}}}}}}$

Let the three consecutive multiples of 9 be k, k + 9 and k + 18.

Given that, Their sum is 81.

According to the Question,

k + k + 9 + k + 18 = 81

⇒ 3k + 27 = 81

⇒ 3k = 81 - 27

⇒ 3k = 54

⇒ k = 18

∴ The Required Smaller number is:
k ⇒ 18 .

And, Second number is:
k + 9 ⇒ 18 + 9 ⇒ 27.

Also, Third number is:
k + 18 ⇒ 18 + 18 ⇒ 36

$\green{\boxed{\pink{\boxed{\blue{\textsf{Hope It Helps !}}}}}}$