Answer 1

Let a be the first term, d be the common difference and T_n be the last of the AP.
Given:
a = 8
T_n = 65
S_n = 730

We know:
T_n = a + (n − 1)d
⇒ 65 = 8 + (n − 1)d
⇒ 57 = (n − 1)d                   ...(i)

and `Sn=n/2(a+Tn)`

`730=n/2(8+65)`

`n/2=730/73`

n=2x10=20

On substituting n = 20 in (i), we get:
57 = (20 − 1)d
⇒ 57 = (19)d
⇒ d = 57/19=3
Thus, the common difference is 3.