Hafizun

Let first term= a second term= T2 third term= T3 nth term= Tn common differnce= d So, a= 6 T2= P T3= 14 d= ? Tn= a + (n - 1)d T2= a + (2 - 1)d= P = a + d= P = 6 + d= P T3= a + (3 - 1)d= 14 = a + 2d= 14 = 6 + 2d= 14 = 2d= 14 - 6 = 2d= 8 = d= 8/2= 4 = d= 4...