Question

the 7th term of an ap is 4 times of second term and 12th term is 2 more than 3 times of its fourth term .find the progression

Answer 1

Solution:

Let the first term of that AP is a,

its common difference is d.

So,its 7th term will be

$a + 6d$

Second term is
$a + d$
A.T.Q.

$(a + 6d) = 4(a + d) \\ \\ a - 4a + 6d - 4d = 0 \\ \\ - 3a + 2d = 0...eq1$
and 12th term is 2 more than 3 times of its fourth term

$(a + 11d) - 2 = 3(a + 3d) \\ \\ a + 11d - 3a - 9d = 2 \\ \\ - 2a + 2d = 2 ...eq2$
Subtract both equations 1 and 2

$- 3a + 2d = 0 \\ \\ - 2a + 2d = 2 \\ + \: \: \: \: \: \: - \: \: \: \: \: \: \: \: - \\ - - - - - - - - \\ - a = - 2 \\ \\ a = 2$
put the value of a in eq 1

$- 3(2) + 2d = 0 \\ \\ 2d = 6 \\ \\ d = 3$
Thus the A.P. is 2,5,8,11...

Hope it helps you.

Answer 2

Given: 7th term is 4 times of 2nd term, 12th term is 2 more than 3 times of its fourth term,

To find: arithmetic progression

Solution:

• Let the first term of AP be x, and the common difference be y.

• Now we can identify the 7th term of an AP:

             x + 6y

             and the second term will be:

             x+ y

• Now, we have given that 7th term is 4 times of 2nd term, so it becomes:

            x + 6y = 4 (x+y)

            x + 6y = 4x + 4y

            2y = 3x................(i)

• Also, we have given that 12th term is 2 more than 3 times of its fourth term, so it becomes:

           x+11y - 2 = 3( x + 3y)

           x+11y - 2 = 3x + 9y

           2y = 2x + 2............(ii)

• Subtracting the equations we get,

            x = 2

• Put x = 2 in equation (i), we get:

            2y = 3(2)

             y = 3

             Formula for AP is x, (x+y), (x+2y), (x+3y), .................

Answer:

           So the obtained AP is 2, 5, 8, 11,................