Question

Find the value of 'd' so that:
(a) (2d + 1), (2d - 1) and (3d + 4) are in AS.​

Answer 1

Given: (2d + 1), (2d - 1) and (3d + 4) are in an AS.

To find: The value of 'd'.

Answer:

In an arithmetic sequence, the difference between every two terms will be the same.

⇒ (2d - 1) - (2d + 1) = (3d + 4) - (2d - 1)

⇒ 2d - 1 - 2d - 1 = 3d + 4 - 2d + 1

⇒ -2 = d + 5

⇒ -2 - 5 = d

⇒ -7 = d

Verification:

To verify, let's find the value of each term by substituting the value of 'd' and verifying the differences.

• (2d + 1) = 2(-7) + 1 = -14 + 1 = -13
• (2d - 1) = 2(-7) - 1 = -14 - 1 = -15
• (3d + 4) = 3(-7) + 4 = -21 + 4 = -17

Verifying every two terms' differences,

-15 - (-13) = -2

-17 - (-15) = -2

Therefore, the value of 'd' is -7.

Answer 2

Answer:

They are in AP then common difference of successive no. is same

therefore ,

Explanation:

3d+4 -(2d-1) = 2d-1(2d+1)

=> 3d+4-2d+1 = 2d-1-2d -1

=> d+5 = -2

=> d = -2-5

=> d = -7

Hence the value of d is -7