The seventh term of an Arithmetic progression is four times its second

1.	The seventh term of an Arithmetic progression is four times its secondterm and twelfth term is 2 more than three times of its fourth term. Findthe progression
Answer» nth term of an APan = a + (n - 1) Then, 7th term of an APa7 = a + 6d 2nd term of an APa2 = a + d 12th term of an APa12 = a + 11d 4th term of an APa4 = a + 3d As per given conditiona + 6d = 4(a + d)3a = 2da = 2d/3.....(1) a + 11d = 3(a + 3d) + 22a - 2d = - 2....(2) Put value of a from eq(1) in eq(2)4d/3 - 2d = - 2(4d - 6d) = - 6d = 6/2 = 3 Hence, a = 2/3 * 3 = 2 Therefore, AP is 2, 5, 8, 11, 14

The seventh term of an Arithmetic progression is four times its secondterm and twelfth term is 2 more than three times of its fourth term. Findthe progression

Answer»

nth term of an APan = a + (n - 1)

Then, 7th term of an APa7 = a + 6d

2nd term of an APa2 = a + d

12th term of an APa12 = a + 11d

4th term of an APa4 = a + 3d

As per given conditiona + 6d = 4(a + d)3a = 2da = 2d/3.....(1)

a + 11d = 3(a + 3d) + 22a - 2d = - 2....(2)

Put value of a from eq(1) in eq(2)4d/3 - 2d = - 2(4d - 6d) = - 6d = 6/2 = 3

Hence, a = 2/3 * 3 = 2

Therefore, AP is 2, 5, 8, 11, 14