Simplify the equation as
(x + 1) + (x + 3) + (x + 5) + … + (x + 59) = 1740
30x + (1 + 3 + 5 + … + 59) = 1740
There are 30 grouped terms because each group is x plus the n-th positive odd integer, which we can write as 2n - 1 for n ≥ 1. Then 2n - 1 = 59 when n = 30.
Let S be the sum of the first n positive integers.
S = 1 + 2 + 3 + … + (n - 2) + (n - 1) + n
Reverse the order of terms in the sum,
S = n + (n - 1) + (n - 2) + … + 3 + 2 + 1
Double S by adding terms in the same position.
2S = (1 + n) + (2 + (n - 1)) + … + ((n - 1) + 2) + (n + 1)
2S = n (n + 1)
S = n (n + 1)/2
Then the sum of the first 2n positive integers is 2n (2n + 1)/2 = 2n² + n.
Let S' be the sum of the first n odd positive integers.
S' = 1 + 3 + 5 + … + (2n - 5) + (2n - 3) + (2n - 1)
We have
2n² + n = 1 + 2 + 3 + 4 + … + (2n - 3) + (2n - 2) + (2n - 1) + 2n
… … … … = [1 + 3 + … + (2n - 3) + (2n - 1)] + [2 + 4 + … + (2n - 2) + 2n]
… … … … = S' + 2 [1 + 2 + … + (n - 1) + n]
… … … … = S' + 2S
⇒ S' = (2n² + n) - (n² + n) = n²
All this to say
30x + (1 + 3 + 5 + … + 59) = 1740
reduces to
30x + 30² = 1740
30x = 840
x = 28