Question

the seventeenth term of an ap is four times its second term and twelth term is 2 more than three 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 17th term will be

$a + 16d$

Second term is
$a + d$

A.T.Q.

$(a + 16d) = 4(a + d) \\ \\ a - 4a + 16d - 4d = 0 \\ \\ - 3a + 12d = 0...eq1$
and 12th term is 2 more than 3 times of its fourth term

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

Multiply eq2 by 3 and

Subtract both equations 1 and 2

$- 3a + 12d = 0 \\ \\ - 3a + 3d = 3 \\ + \: \: \: \: \: \: - \: \: \: \: \: \: - \\ - - - - - - - \\ 9d = - 3 \\ \\ d = = \frac{ - 3}{9} \\ \\ d = \frac{ - 1}{3}$

put the value of d in eq2

$- a - \frac{1}{3} = 1 \\ \\ - a = 1 + \frac{1}{3} \\ \\ a = - \frac{4}{3}$

So, the AP is

$\frac{ - 4}{3},\: \frac{ - 5}{3}, \frac{ - 6}{3}, ...$

Answer 2

Answer:

The required A.P is

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

Step-by-step explanation:

I think your question must be

"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."

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........................