Answer :

Given, first term a = 1, last term l = 11 and S_{n} = 36.

Since, n^{th} term a_{n} is given by :

a_{n} = a + (n - 1)d

Where,

a = First term of AP

d = Common difference of AP

and no of terms is ‘n’

⇒ 11 = 1 + (n - 1)d

⇒ 10 = (n -1)d … (i)

Since the sum of n terms is

S_{n =} _{}

⇒ 72 = n[ 2 + (n-1)d] …(ii)

From eq. (i) and (ii), we get,

⇒ 72 = n[ 2 + 10]

⇒ 72 = 12n

⇒ n = 6

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