Find x so that, x²+3x+1,x^2 +5x + 16 and x2 -7x + 3 are the consecutive terms of AP

2 thoughts on “Find x so that, x²+3x+1,x^2 +5x + 16 and x2 -7x + 3 are the consecutive terms of AP<br />”

1. Solution –

   Firstly, we have three equations

   • x² + 3x + 1
   • x² + 5x + 16
   • x² – 7x + 3

   ⠀

   Now, we have to find the value of x so that these equations are the consecutive terms of A.P.

   Since, in an A.P. a, b and c

   ⠀⠀⠀⠀⠀⠀❏ 2b = a + c

   ⠀

   Consider,

   • a = x² + 3x + 1
   • b = x² + 5x + 16
   • c = x² – 7x + 3

   ⠀

   Substituting the values

   → 2(x² + 5x + 16) = x² + 3x + 1 + x² – 7x + 3

   ⠀

   → 2x² + 10x + 32 = 2x² – 4x + 4

   ⠀

   Cancelling 2x² both the sides

   → 10x + 32 = -4x + 4

   ⠀

   → 10x + 4x = 4 – 32

   ⠀

   → 14x = -28

   ⠀

   → x = -28/14

   ⠀

   → x = –2

   ⠀

   Hence,

   • Required value of x is -2.

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2. Answer:–

   Given:

   x² + 3x + 1 , x² + 5x + 16 , x² – 7x + 3 are consecutive terms of an AP.

   We know that;

   If a , b , c are in AP then,

   ⟹ 2b = a + c

   Let;

   • a = x² + 3x + 1
   • b = x² + 5x + 16
   • c = x² – 7x + 3.

   According to the question;

   ⟹ 2(x² + 5x + 16) = x² + 3x + 1 + x² – 7x + 3

   ⟹ 2x² + 10x + 32 = x² + x² + 3x – 7x + 1 + 3.

   ⟹ 2x² + 10x + 32 = 2x² – 4x + 4

   ⟹ 2x² + 10x – 2x² + 4x = 4 – 32

   ⟹ 14x = – 28

   ⟹ x = – 28/14

   ⟹ x = – 2

   ∴ The value of x is – 2.

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