Let first term of AP = a

     Last term of AP = l

    common difference = d

    number of terms in AP = n

 nth term in AP = a + (n-1)d

So 22th term = a + (22-1)d = a + 21*d        (Given n =22)

Given that 

               22th term = 149

        =>  a + 21*d = 149

        => a + 21*7 = 149     (Given d = 7)

        => a + 147 = 149

        => a  = 149 - 142

        => a = 2

Now sum of AP = (n/2)*{2a + (n-1)*d} 

=> (22/2)*{2*2 + (22-1)*7}           

=> 11*(4 + 21*7)

=> 11*(4 + 147)

=> 11*151

=> 1661

So first terms in AP is 2 and sum is 1661