1+5+9+----+x

This is an arithmetic progression with common difference = 4. We have to find out the nth term for which the sum of the series upto that term becomes equal to 231.

We know, the sum of an arithmetic progression is given by

`S = n/2 (2a+(n-1)d) `

Where, a is the first term in the series, n, the total number of terms and d, the common difference.

Putting the values, here

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

Or, `n(1+2n-2)=231`

Or, `n(2n-1)=231`

Or, `2n^2-n-231=0`

`n =[1+-sqrt(1-4*2*(-231))]/(2*2)`

`= (1+ 43)/4 ` and `(1-43)/4`

n is the number of terms in the arithmetic series which must be a positive integer. So the second root of the equation is discarded.

Therefore `n = 44/4 = 11.`

The last term, x is given by

`x = a+ (n-1)d`

`=1+(11-1)4`

`=1+40`

`=41`

**Therefore, the value of x in the given series is 41**.

`x=(n-1)4 +1= 4n-3`

`S_n=(1+x)/2 xx n=(1+4n-3)/2 xxn=(2n-1)n`

`231=2n^2-n`

`2n^2-n-231=0`

`Delta= 1-4xx2xx(-231)=1849`

`n=(1+- 43)/4` `n=11` `n=-21/2` (discarged for negative and not integer )

then `x= 1 +(11-1)xx 4=41`