Question

34.
The seventh term of an Arithmetic progression is four times its second
term and twelfth term is more than three times of its fourth term. Find
3
the progression​

Answer 1

a7 = 4 × a2

a + 6d = 4 × (a + d)

a + 6d = 4a + 4d

6d - 4d = 4a - a

2d = 3a -> ( 1 )

a12 = 2 + 3 × (a4)

a + 11d = 2 + 3 × (a + 3d)

a + 11d = 2 + 3a + 9d

11d - 9d = 2 + 3a - a

2d = 2 + 2a

From eq. ( 1 )

3a = 2 + 2a

3a - 2a = 2

a = 2-> ( 2 )

Putting value of eq. ( 2 ) in eq. ( 1 )

2d = 3a

2d = 3(2)

2d = 6

d = 6/2 = 3

Hence, the A.P. is

a, a+d, a+2d, a+3d..............a+(n-1)d

2, 5, 8, 11..............so on.

Hope that it is helpful to you.

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