(a) \(T_{n} = a + (n - 1)d\) (For an AP series)
\(T_{6} = 37 = a + (6 - 1) d = a + 5d\)
\(a + 5d = 37 .... (1)\)
\(S_{n} = \frac{n}{2}(2a + (n - 1) d)\)
or
\(S_{n} = \frac{n}{2}(a + l)\) where a = first term; and l = last term.
Since, we are given the sum of the first 6 terms and also given the sixth term which is the last term in the sum, we can use the second formula where the sixth term is the last term in this case.
\(S_{6} = 147 = \frac{6}{2} (a + 37)\)
\(147 = 3(a + 37) \implies 147 = 3a + 111 \)
\(3a = 147 - 111 = 36 \implies a = 12\)
(b) From equation (1) above,
\(a + 5d = 37\)
\(12 + 5d = 37\)
\(5d = 37 - 12 = 25\)
\(d = 5\)
\(S_{15} = \frac{15}{2} (2(12) + (15 - 1)(5))\)
\(\frac{15}{2} (24 + 70) = \frac{15}{2} (94)\)
= \(15 \times 47\)
= \(705\).