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

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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

shaznl1's profile pic

shaznl1 | High School Teacher | (Level 1) Salutatorian

Posted on

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 

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