# What is the natural number x if 1+ 5 + 9 + ...+ x = 231?

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The series given to us 1, 5, 9... has the same difference between consecutive terms which is 4. This makes it an arithmetic progression.

Let n be the nth term of the series. The sum of n terms of an AP is given by (2a + (n-1)d)(n/2)

=> (2 + (n - 1)*4)(n/2) = 231

=> n + n(n - 1)*2 = 231

=> n + 2n^2 - 2n = 231

=> 2n^2 - n - 231 = 0

=> 2n^2 + 21n - 22n - 231 = 0

=> 2n(n + 10.5) - 22(n + 10.5) = 0

=> (2n - 22)(n + 10.5) = 0

=> n = 11 and n = -10.5

We can eliminate the negative root.

x = 1 + (11 - 1)*4

=> 1 + 40

=> 41

**The value of x = 41**

The series in this question is called a arithmetic sequence. The difference between each number is 4

The first number is 1, the last number is X, there are (x-1)/4 +1 terms in the sequence (you can verify that using a lot of ways)

then the total sum is

((1+X)*（（x-1)/4+1))/2 =231

(x^2-1)/4 +1+X=462 (multiplication Property of Equality(P.O.E))

x^2 +3 +4X = 1848 ( Addition and Multiplication P.O.E)

X^2 +4X - 1845 =0 (Simplify)

(x+45)(x-41) = 0 (Factoring)

x= -45 or 41 Eliminate the negative root

The Term is 41