The third term of an A.P is 6 and the fifth term is 12. Find the sum of its first twelve terms
A.
201
B.
144
C.
198
D.
72
Correct answer is C
T\(_3\) = 6
T\(_5\) = 12
S\(_{12}\) = ?
T\(_n\) = a + (n - 1)d
⇒ T\(_3\) = a + 2d = 6 ----- (i)
⇒ T\(_5\) = a + 4d = 12 ----- (ii)
Subtract equation (ii) from (i)
⇒ -2d = -6
⇒ d\(\frac{-6}{-2}\) = 3
Substitute 3 for d in equation (i)
⇒ a + 2(3) = 6
⇒ a + 6 = 6
⇒ a = 6 - 6 = 0
S\(_n\) = \(\frac{n(2a + (n - 1)d)}{2}\)
⇒ S\(_{12}\) = \(\frac{12(2 \times 0 + (12 - 1)3)}{2}\)
⇒ S\(_{12}\) = 6(0 + 11 x 3)
⇒ S\(_{12}\) = 6(33)
∴ S\(_{12}\) = 198
