The sum of the first twelve terms of an Arithmetic Progression is 168. If the third term is 7, find the values of the common difference and the first term.
A.P ; \(S_{12} = 168 ; T_{3} = 7\)
\(S_{n} = \frac{n}{2}(2a + (n - 1) d)\)
\(T_{n} = a + (n - 1)d\)
\(168 = \frac{12}{2}(2a + (12 - 1)d \implies 168 = 6(2a + 11d)\)
\(168 = 12a + 66d \implies 84 = 6a + 33d ..... (1)\)
\(7 = a + 2d .... (2)\)
From (2), a = 7 - 2d. Put into (1), we have
\(84 = 6(7 - 2d) + 33d \)
\(84 = 42 - 12d + 33d \implies 84 - 42 = 21d\)
\(21d = 42 \implies d = 2\)
\(a = 7 - 2(2) = 7 - 4 = 3\)
\(a(\text{first term}) = 3 ; d(\text{common difference}) = 2\)
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