Three times the smallest of three consecutive odd numbers decreased by $7$ equals twice the larger one. Find the numbers.

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Solution

Let odd numbers are
$2 n + 1, 2 n + 3, 2 n + 5$
Now according to question
${3 (2 n + 1) - 7} = 2 {2 n + 5}$
$6 n + 3 - 7 = 4 n + 10$
$2 n = 14 \Rightarrow n = 7$
So numbers are $2 \times 7 + 1, 2 \times 7 + 3, 2 \times 7 + 5$
$15, 17, 19$

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Three times the smallest of three consecutive odd numbers decreased by 7 equals twice the larger one. Find the numbers.

Let odd numbers are 2n+1,2n+3,2n+5Now according to question{3(2n+1)−7}=2{2n+5}6n+3−7=4n+102n=14⇒n=7So numbers are 2×7+1,2×7+3,2×7+515,17,19

Three times the smallest of three consecutive odd numbers decreased by $7$ equals twice the larger one. Find the numbers.

Let odd numbers are
$2 n + 1, 2 n + 3, 2 n + 5$
Now according to question
${3 (2 n + 1) - 7} = 2 {2 n + 5}$
$6 n + 3 - 7 = 4 n + 10$
$2 n = 14 \Rightarrow n = 7$
So numbers are $2 \times 7 + 1, 2 \times 7 + 3, 2 \times 7 + 5$
$15, 17, 19$