Question

In an arithmetic progression,there are 6 terms and sum is 3,the first term is 4 times the third term.the fifth term in the progression is?

Answer 1

Let first term be 'a' and difference between the two successive terms be 'd'.  Sum S = n/2(2a + (n-1) d)  3 = 6/2(2a + (6-1)d) 2a+5d=1......eqn 1  1st term =4*(3rd term) a = 4*(a + 2d) 3a + 12d = 0.....eqn 2  On solving eqn 1 and 2, 6a + 15d =3 6a + 24d =0  -3 = 9d  d=-1/3  2a + 5(-1/3) = 1 2a = 1 + 5/3 a=4/3  Fifth term = a + 4d                  = 4/3 + 4(-1/3) =0