(1)7, 9, 1 1, ..., 49 form a arithmetic progression with the first term of 7 and the tolerance of 2.

Let the general formula of this arithmetic progression be an = 7+2 (n- 1) = 2n+5.

So 2n+5 = 49.

Then n=22

(2)7+9+ 1 1+……+49

=22*(7+49)/2

= 1 1*56

=6 16