Question

if the sum of n terms of an ap is n^2 then find the common difference?​

Answer 1

Given:

Sn=n² ---(1)

We know

Sn=(n/2)[2a+(n-1)d] ---(2)

using (1) in (2) we get

n²=(n/2)[2a+(n-1)d]

2n²=2an + n²d -nd

(2-d)n²=(2a-d)n

2n-nd=2a-d

2n-2a=nd-d

2(n-a)=(n-1)d

Hence, d=2(n-a)/(n-1)

Check it out dear friend....

Answer 2

Answer:

2 (n - a) / (n - 1)

Step-by-step explanation:

Let t1 = a c.d. = d [d not equals

to 0]

Sn = n/2 [ 2 t1 + (n - 1) d]

=> n^2 = n/2 [ 2a + (n - 1) d]

=> 2n = 2a + (n - 1) d

Therefore , d = 2 (n - a) / (n - 1) [ANS.]

HOPE THIS WILL HELP YOU !

THANK YOU.