The third term of an A.P is 6 and the fifth term is 12. Find

Explanation

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