If 12th term of an A.P. is 13 and the sum of the first four terms is 24, what is the sum of first 10 terms?
a12= -13
a + (12 – 1) d = -13
a + 11d = -13
S4 = 
[2a + (4 – 1) d]
24 = 2 [2a + 3d]
12 = 2a + 3d
12 = 2 (-11d – 13) + 3d
12 = -22d – 26 + 3d
38 = -19d
d = -2
a = -13 + 11(2)
= -13 + 22
= 9
S10= 
[2(9) + (10 – 1) (-2)]
= 5 [18 – 18]
= 0
9