If the sum of the first four terms of an A.P. is 6 and the sum of its first six terms is 4, then the sum of its first twelve terms is
- -26
- -20
- -24
- -22
Answer
(4) -22
Solution
Let the first term be a and the common difference d, so Sn = n2[2a + (n-1)d].
S4 = 2(2a + 3d) = 6
2a + 3d = 3 ...(i)
S6 = 3(2a + 5d) = 4
2a + 5d = 43 ...(ii)
Subtracting (i) from (ii), 2d = 43 - 3 = - 53, so d = - 56, and then from (i), 2a = 3 + 52 = 112.
S12 = 6(2a + 11d) = 6 ( 112 - 556 ) = 33 - 55 = -22
Answer: (4)