Question

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

Answer 1

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Assumption

First term = p

Common difference = d

${\boxed{\sf\:{Seventh\;Term}}}$

= p + 6d

${\boxed{\sf\:{Second\;Term}}}$

= p + d

$\large{\boxed{\sf\:{Seventh\;Term=4\times Second\;Term}}}$

p + 6d = 4 (p + d)

p + 6d = 4p + 4d

2d = 3p ............. (1)

${\boxed{\sf\:{12^{th} \;Term}}}$

= p + 11d

${\boxed{\sf\:{4^{th} \;Term}}}$

= p + 3d

${\boxed{\sf\:{12^{th} \;Term}}}$

= 2 + 3 × 4th term

p + 11d = 2 + 3 × (p + 3d)

p + 11d = 2 + 3p + 9d

2d = 2p + 2 ......... (2)

$\large{\boxed{\sf\:{Solving\;Equation\;we\;get}}}$

3p = 2p + 2

p = 2

2d = 3p

2d = 6

$\tt{\rightarrow d=\dfrac{6}{2}}$

d = 3

${\boxed{\sf\:{Arithmetic\; Progression=2+(n-1)3}}}$

= 2 + 3n - 3

= 3n - 1

$\Large{\boxed{\sf\:{Arithmetic\;Progression=2,5,8,11}}}$

Answer 2

SOLUTION:-

According to the question:

Let the First term of an A.P. is a,

Common difference is d

Therefore,

7th term of an AP;

${}^{a} 7 = a + (7 - 1)d \\ \\ = > a + 6d$

2nd term of an A.P;

${}^{a} 2 = a + (2 - 1)d \\ \\ = > a + d$

So,

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

=) a+ 6d = 4a + 4d

=) a-4a +6d -4d =0

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

Now,

12th term of an AP;

${}^{a} 12 = a + (12 - 1)d \\ \\ = > a + 11d$

In this AP 2 is more than 3 times of its fourth term;

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

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

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

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

Subtracting equation (1) & (2) we get;

=) -a = -2

=) a= 2

Putting the value of a in equation (1) we get;

=) -3(2) + 2d=0

=) -6 + 2d = 0

=) 2d = 6

=) d= 6/2

=) d= 3

Hence,

The A.P. is 2,5,8,11....

Hope it helps ☺️