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

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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**

1+5+9+...+n = 231.

To find n.

T1 = 1.

T2 = 5

T3 = 9.

T3-T2= T2-T1 = 4.

Therefore Tr = 1+4(r-1) = 4r-3.

Therefore sum to n terms Sm = (T1+Tm)m/2 = (1+4m-3)m/2.

Let Sm = (4m-2)m/2 = 231.

(2m-1)m = 231.

2m^2-m - 231 = 0.

(2m +21)(m-11) = 0.

2m +21 = 0 , or m-11 =0.

m = -21/2 . Or m = 11.

We ignore m = -21/2 as it is a fraction and negative.

m = 11 .

So Tm = 11.

Tm = 1+4*(11-1) = 41.

Therefore 1+5+9+....+41 = **231**.