Series: a, a+d, a+2d, ..., a+(n-1)d up to the nth term.
Sum Sn of first n terms can be found by taking pairs of terms:
(a+a+(n-1)d)+(a+d+a+(n-2)d)+...
The first pair is the sum of the first and last terms: 2a+(n-1)d.
The second pair is the sum of the 2nd term a+d and the penultimate term: 2a+(n-1)d.
So all the pairs have the same sum: 2a+(n-1)d, and because there are n terms in the series there must be n/2 pairs.
Therefore Sn=(n/2)(2a+(n-1)d)=an+dn(n-1)/2.
S29=29a+d(29)(28)/2=29a+406d=1102.
Divide through by 29: a+14d=38 (proves (a))
(b) 2nd term is a+d and the 7th term is a+6d. The sum is 2a+7d=13.
We know that a+14d=38, so a=38-14d. Substitute for a in 2a+7d=13: 76-28d+7d=13, 63=21d, d=63/21=3, so a=38-42=-4.
Therefore a=-4 and d=3.