Question

Q10. If the numbers 2n-1, 3n+2 and 6n - 1 are in A.P, find n and hence find the numbers.​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

The value of n is 3.

The numbers in the AP are 5, 11 & 17.

Step-by-step-explanation:

We have given that,

The numbers 2n - 1, 3n + 2 and 6n - 1 are in AP.

We have to find the value of n and the numbers.

Here,

• t₁ = a = 2n - 1
• t₂ = 3n + 2
• t₃ = 6n - 1

Now, we know that,

The common difference between two consecutive terms of an A.P. is constant.

∴ d = t₂ - t₁ = t₃ - t₂

⇒ t₂ - t₁ = t₃ - t₂

⇒ 3n + 2 - ( 2n - 1 ) = 6n - 1 - ( 3n + 2 )

⇒ 3n + 2 - 2n + 1 = 6n - 1 - 3n - 2

⇒ 3n - 2n + 2 + 1 = 6n - 3n - 1 - 2

⇒ n + 3 = 3n - 3

⇒ 3 + 3 = 3n - n

⇒ 6 = 2n

⇒ n = 6 ÷ 2

⇒ n = 3

∴ The value of n is 3.

Now,

The first number in AP = 2n - 1

⇒ The first number in AP = 2 * 3 - 1

⇒ The first number in AP = 6 - 1

⇒ The first number in AP = 5

Now,

The second number in AP = 3n + 2

⇒ The second number in AP = 3 * 3 + 2

⇒ The second number in AP = 9 + 2

⇒ The second number in AP = 11

Now,

The third number in AP = 6n - 1

⇒ The third number in AP = 6 * 3 - 1

⇒ The third number in AP = 18 - 1

⇒ The third number in AP = 17

∴ The numbers in the AP are 5, 11 & 17.