Question

the 7th term of an arithmetic progression is four times its second term and twelfth term is 2 more than three times of its fourth term.Find progression.
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Answer 1

Answer:

The required A.P is

2, 5, 8, 11........................

Step-by-step explanation:

Formula used:

The n th term of the A.P a, a+d, a+2d, .......

is

$t_n=a+(n-1)d$

$Given:t_7=4\:t_2\\\\a+6d=4[a+d]\\\\3a-2d=0.........(1)$

$Also,\\\\t_12=3\:t_4+2\\\\a+11d=3(a+3d)+2\\\\a+11d=3a+9d+2\\\\2a-2d=-2.........(2)$

Now we solve (1) and (2)

3a-2d=0.........(1)

2a-2d=-2.........(2)

(1)-(2)

a=2

using a=2 in (1) we get

3(2)-2d=0

2d=6

d=3

The required A.P is

2, 5, 8, 11........................

Answer 2

Answer:

3n - 1

2 , 5 , 8 , 11 ......

Step-by-step explanation:

the 7th term of an arithmetic progression is four times its second term and twelfth term is 2 more than three times of its fourth term.Find progression.

7th Term = a + 6d

2nd term = a + d

7th term = 4 * (2nd Term)

=> a + 6d = 4 (a + d)

=> a + 6d = 4a + 4d

=> 2d = 3a   - eq 1

12th term = a + 11d

4th term = a + 3d

12th term = 2 + 3 * 4th term

a + 11d = 2 + 3 *(a + 3d)

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

=> 2d = 2a + 2   - eq2

Equating both equations

3a = 2a + 2

=> a = 2

2d =3a

=> 2d = 6

=> d= 3

AP = 2 + (n-1)3

= 2 + 3n - 3

= 3n - 1

2 , 5 , 8 , 11 ......