Question

The 7th term of an AP is 4 times its 2nd term is 2more than 3 times of its 4th term. Find the progression .​

Answer 1

Correct Question :-

The 7th term of an AP is 4 times its 2nd term and it's 12th term is 2 more than 3 times of its 4th term. Find the progression

Solution :-

According to the question ,

7th term of an AP = 4 ( 2n term of an AP)

Subsitute the required values,

a + 6d = 4 ( a + d)

a + 6d = 4a + 4d

6d - 4d = 4a - a

2d = 3a

a = 2d/3. ...eq( 1 )

Now,

It's 12th term is 2 more than 3 times of its 4th term.

Therefore,

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

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

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

-2a = -2d + 2

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

Subsitute eq( 1) in eq( 2 )

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

-4d/3 + 2d = 2

-4d + 6d = 6

2d = 6

d = 6/2 = 3

Thus, The value of d is 3

Subsitute value of d in eq ( 1 )

a = 2 * 3/3

a = 2

So, The arithmetic progression are :-

2 , 4 , 6 , 8 ...$.$

Answer 2

Question:

• The 7th term of an AP is four times its 2nd term and its 12th term is 2 more than three times of its 4th term. Find the progression.

Answer:

• The progression is: 2, 5, 8, 11, 14, 17...

Step-by-step explanation:

Given that:

• The 7th term of an AP is four times its 2nd term.
• 12th term is 2 more than three times of its 4th term.

To Find:  

• The progression.

Let us assume:  

• First term of an AP = a
• Common difference = d

We know that:

• aₙ = a + (n - 1)d

Where,

• aₙ = nth term
• a = First term
• n = Number of terms
• d = Common difference

According to the question:

The 7th term of an AP is four times its 2nd term.

⟶ a₇ = 4 {a₂}

⟶ a + (7 - 1)d = 4 {a + (2 - 1)d}

⟶ a + 6d = 4 {a + d}

⟶ a + 6d = 4a + 4d

⟶ 6d - 4d = 4a - a

⟶ 2d = 3a _____(i)

12th term is 2 more than three times of its 4th term.

⟶ a₁₂ = 3 {a₄} + 2

⟶ a + (12 - 1)d = 3 {a + (4 - 1)d} + 2

⟶ a + 11d = 3 {a + 3d} + 2

⟶ a + 11d = 3a + 9d + 2

⟶ 11d - 9d = 3a - a + 2  

⟶ 2d = 2a + 2

⟶ 3a = 2a + 2 [from eqⁿ (i): 2d = 3a]

⟶ 3a - 2a = 2

⟶ a = 2

∴ First term = 2

In equation (i),

⟶ 2d = 3a

⟶ 2d = 3(2) [The value of a = 2]

⟶ 2d = 6

⟶ d = 6/2

⟶ d = 3

∴ Common difference = 3

Some of the terms of AP:

• a₁ = a = 2
• a₂ = a₁ + d = 2 + 3 = 5
• a₃ = a₂ + d = 5 + 3 = 8
• a₄ = a₃ + d = 8 + 3 = 11
• a₅ = a₄ + d = 11 + 3 = 14
• a₆ = a₅ + d = 14 + 3 = 17

Therefore,

• The progression is: 2, 5, 8, 11, 14, 17...

Hope it helps you!!(●'◡'●)