Question

Find three consecutive numbers such that seven times the smallest number may be equal to three times the sum of other two

Answer 1

let the numbers be a-1,a and a+1
ATQ,
$7(a - 1) = 3(a + a + 1) \\ 7a - 7 = 6a + 6 \\ a = 13$
therefore numbers are 12,13,14

Answer 2

$\huge\mathfrak\red{Answer :}$


Let three consecutive numbers be x, x + 1, x + 2.

Seven times the smallest term....means 7x

is equal to three times the sum of others two....means 3(x + 1 + x + 2)

So,

7x = 3(x + 1 + x + 2)

7x = 3(2x + 3)

7x = 6x + 9

7x - 6x = 9

x = 9

x + 1 = 9 + 1 = 10

x + 2 = 9 + 2 = 11


So three consecutive numbers are.... $\textbf{9, 10, 11}$

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