Find the natural number n if 1+5+9+..+n=231.

The series given is an arithmetic progression with the
first term 1 and common difference as 4.

Now the sum of the
first N terms of an AP is given by (N/2)[2a1 +
d(N-1)]

=(N/2)[ 2 +
(N-1)*4]

This is equal to
231

So (N/2)[ 2 + (N-1)*4] =
231

=> N[ 2 + (N-1)*4] =
462

=> 2N + 4*N*(N-1) =
462

=> 2N + 4N^2 - 4N =
462

=> 4N^2 - 2N - 462 =
0

=> 2N^2 - N -231 =
0

Therefore the roots are N = 11 and N =
10.5

We only consider the value N =
11.

So n = 1 + (N-1)*4 = 1 + 10*4 =
41

Therefore the term n required is 1 + 10*4
= 41