Question

the seventh 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 the progression ​

Answer 1

A.T.Q.

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.

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Answer 2

The arithmetic progression are 2, 5, 8, .... .

Step-by-step explanation:

Let first term = a and common difference = d

Find the progression ​= ?

According to question,

$a_{7} =4\times a_{2}$

⇒$a+6d=4\times (a+d)$

⇒ $a+6d=4a+4d$

⇒ $3a-2d=0$         .....(1)

Also,

$a_{12} =3\times a_{4}+2$

⇒$a+11d=3(a+3d)+2$

⇒ $a+11d=3a+9d+2$

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

Subtracting (1) from (2), we get

a = 2

Put a = 2 in (1), we get

$3(2)-2d=0$

∴ d = 3

The arithmetic progression are

2, 2 + 3 , 2 + 2 × 3, ...

= 2, 5, 8, ....

Hence, the arithmetic progression are 2, 5, 8, .... .