Question

2x+1,4x-1,5x+1 are in the arithmatic sequence . find the value of x .write its nth term .find the position of 195​

Answer 1

In an AP,

→ a₂ - a₁ = a₃ - a₂ = d

⇒4x - 1 - (2x + 1) = 5x + 1 - (4x - 1)

⇒4x - 1 - 2x - 1 = 5x + 1 - 4x + 1

⇒2x - 2 = x + 2

⇒2x - x = 2 + 2

⇒x = 4

By substituting the value of x in a₁ :-

2x + 1 = 2(4) + 1

⇒a₁ = 9

Common difference, d = a₂ - a₁

⇒d = 4(4) - 1 - a₁

⇒d = 15 - 9

⇒d = 6

Now,Let 195 be the nth term of the AP.

We know,

aₙ = a + (n - 1)d

Substituting the values :-

195 = 9 + (n - 1)6

→195 - 9 = 6n - 6

→186 + 6 = 6n

→192 = 6n

→n = 192/6

→n = 32

Therefore, 195 is the 32nd term of the AP.

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